class Address {
int i ;
char b;
string c;
public:
void showMap ( void ) ;
};
void Address :: showMap ( void ) {
cout << "address of int :" << &i << endl ;
cout << "address of char :" << &b << endl ;
cout << "address of string :" << &c << endl ;
}
The output is:
address of int : something
address of char : // nothing, blank area, that is nothing displayed
address of string : something
Why?
Another interesting thing: if int, char, string is in public, then the output is
... int : something
... char :
... string : something_2
something_2 - something is always equal to 8. Why? (not 9)
When you are taking the address of b, you get char *. operator<< interprets that as a C string, and tries to print a character sequence instead of its address.
try cout << "address of char :" << (void *) &b << endl instead.
[EDIT] Like Tomek commented, a more proper cast to use in this case is static_cast, which is a safer alternative. Here is a version that uses it instead of the C-style cast:
cout << "address of char :" << static_cast<void *>(&b) << endl;
There are 2 questions:
Why it does not print the address for the char:
Printing pointers will print the address for the int*and the string* but will not print the contents for char* as there is a special overload in operator<<. If you want the address then use: static_cast<const void *>(&c);
Why the address difference between the int and the string is 8
On your platform sizeof(int) is 4 and sizeof(char) is 1 so you really should ask why 8 not 5. The reason is that string is aligned on a 4-byte boundary. Machines work with words rather than bytes, and work faster if words are not therefore "split" a few bytes here and a few bytes there. This is called alignment
Your system probably aligns to 4-byte boundaries. If you had a 64-bit system with 64-bit integers the difference would be 16.
(Note: 64-bit system generally refers to the size of a pointer, not an int. So a 64-bit system with a 4-byte int would still have a difference of 8 as 4+1 = 5 but rounds up to 8. If sizeof(int) is 8 then 8+1 = 9 but this rounds up to 16)
When you stream the address of a char to an ostream, it interprets that as being the address of the first character of an ASCIIZ "C-style" string, and tries to print the presumed string. You don't have a NUL terminator, so the output will keep trying to read from memory until it happens to find one or the OS shuts it down for trying to read from an invalid address. All the garbage it scans over will be sent to your output.
You can probably get it to display the address you want by casting it, as in (void*)&b.
Re the offsets into the structure: you observed the string is placed at offset 8. This is probably because you have 32-bit ints, then an 8-bit char, then the compiler chooses to insert 3 more 8-bit chars so that the string object will be aligned at a 32-bit word boundary. Many CPUs/memory-architectures need pointers, ints etc. to be on word-size boundaries to perform efficient operations on them, and would otherwise have to do many more operations to read and combine multiple values from memory before being able to use the values in an operation. Depending on your system, it may be that every class object needs to start on a word boundary, or it may be that std::string in particular starts with a size_t, pointer or other type that requires such alignment.
Because when you pass a char* to std::ostream it will print the C-style (ie: char array, char*) string it points to.
Remember that "hello" is a char*.
The address of char is being treated as a nul-terminated string and is displaying the contents of that address, which is probably undefined, but in this case an empty string. If you cast the pointers to void *, you will get the results you desire.
The difference between something2 and something being 8 is due to aligned and ability of the compiler to decide for itself where in the stack the variables are declared.
For the second issue - the compiler by default will pad structure members. The default pad is to the sizeof(int), 4 bytes (on most architectures). This is why an int followed by a char will take 8 bytes in the structure, so the string member is at offset 8.
To disable padding, use #pragma pack(x), where x is the pad size in bytes.
Your syntax should be
cout << (void*) &b
hrnt is right about the reason for the blank: &b has type char*, and so gets printed as a string until the first zero byte. Presumably b is 0. If you set b to, say, 'A', then you should expect the printout to be a string starting with 'A' and continuing with garbage until the next zero byte. Use static_cast<void*>(&b) to print it as a an address.
For your second question, &c - &i is 8, because the size of an int is 4, the char is 1, and the string starts at the next 8-byte boundary (you are probably on a 64-bit system). Each type has a particular alignment, and C++ aligns the fields in the struct according to it, adding padding appropriately. (The rule of thumb is that a primitive field of size N is aligned to a multiple of N.) In particular you can add 3 more char fields after b without affecting the address &c.
Related
class Address {
int i ;
char b;
string c;
public:
void showMap ( void ) ;
};
void Address :: showMap ( void ) {
cout << "address of int :" << &i << endl ;
cout << "address of char :" << &b << endl ;
cout << "address of string :" << &c << endl ;
}
The output is:
address of int : something
address of char : // nothing, blank area, that is nothing displayed
address of string : something
Why?
Another interesting thing: if int, char, string is in public, then the output is
... int : something
... char :
... string : something_2
something_2 - something is always equal to 8. Why? (not 9)
When you are taking the address of b, you get char *. operator<< interprets that as a C string, and tries to print a character sequence instead of its address.
try cout << "address of char :" << (void *) &b << endl instead.
[EDIT] Like Tomek commented, a more proper cast to use in this case is static_cast, which is a safer alternative. Here is a version that uses it instead of the C-style cast:
cout << "address of char :" << static_cast<void *>(&b) << endl;
There are 2 questions:
Why it does not print the address for the char:
Printing pointers will print the address for the int*and the string* but will not print the contents for char* as there is a special overload in operator<<. If you want the address then use: static_cast<const void *>(&c);
Why the address difference between the int and the string is 8
On your platform sizeof(int) is 4 and sizeof(char) is 1 so you really should ask why 8 not 5. The reason is that string is aligned on a 4-byte boundary. Machines work with words rather than bytes, and work faster if words are not therefore "split" a few bytes here and a few bytes there. This is called alignment
Your system probably aligns to 4-byte boundaries. If you had a 64-bit system with 64-bit integers the difference would be 16.
(Note: 64-bit system generally refers to the size of a pointer, not an int. So a 64-bit system with a 4-byte int would still have a difference of 8 as 4+1 = 5 but rounds up to 8. If sizeof(int) is 8 then 8+1 = 9 but this rounds up to 16)
When you stream the address of a char to an ostream, it interprets that as being the address of the first character of an ASCIIZ "C-style" string, and tries to print the presumed string. You don't have a NUL terminator, so the output will keep trying to read from memory until it happens to find one or the OS shuts it down for trying to read from an invalid address. All the garbage it scans over will be sent to your output.
You can probably get it to display the address you want by casting it, as in (void*)&b.
Re the offsets into the structure: you observed the string is placed at offset 8. This is probably because you have 32-bit ints, then an 8-bit char, then the compiler chooses to insert 3 more 8-bit chars so that the string object will be aligned at a 32-bit word boundary. Many CPUs/memory-architectures need pointers, ints etc. to be on word-size boundaries to perform efficient operations on them, and would otherwise have to do many more operations to read and combine multiple values from memory before being able to use the values in an operation. Depending on your system, it may be that every class object needs to start on a word boundary, or it may be that std::string in particular starts with a size_t, pointer or other type that requires such alignment.
Because when you pass a char* to std::ostream it will print the C-style (ie: char array, char*) string it points to.
Remember that "hello" is a char*.
The address of char is being treated as a nul-terminated string and is displaying the contents of that address, which is probably undefined, but in this case an empty string. If you cast the pointers to void *, you will get the results you desire.
The difference between something2 and something being 8 is due to aligned and ability of the compiler to decide for itself where in the stack the variables are declared.
For the second issue - the compiler by default will pad structure members. The default pad is to the sizeof(int), 4 bytes (on most architectures). This is why an int followed by a char will take 8 bytes in the structure, so the string member is at offset 8.
To disable padding, use #pragma pack(x), where x is the pad size in bytes.
Your syntax should be
cout << (void*) &b
hrnt is right about the reason for the blank: &b has type char*, and so gets printed as a string until the first zero byte. Presumably b is 0. If you set b to, say, 'A', then you should expect the printout to be a string starting with 'A' and continuing with garbage until the next zero byte. Use static_cast<void*>(&b) to print it as a an address.
For your second question, &c - &i is 8, because the size of an int is 4, the char is 1, and the string starts at the next 8-byte boundary (you are probably on a 64-bit system). Each type has a particular alignment, and C++ aligns the fields in the struct according to it, adding padding appropriately. (The rule of thumb is that a primitive field of size N is aligned to a multiple of N.) In particular you can add 3 more char fields after b without affecting the address &c.
class Address {
int i ;
char b;
string c;
public:
void showMap ( void ) ;
};
void Address :: showMap ( void ) {
cout << "address of int :" << &i << endl ;
cout << "address of char :" << &b << endl ;
cout << "address of string :" << &c << endl ;
}
The output is:
address of int : something
address of char : // nothing, blank area, that is nothing displayed
address of string : something
Why?
Another interesting thing: if int, char, string is in public, then the output is
... int : something
... char :
... string : something_2
something_2 - something is always equal to 8. Why? (not 9)
When you are taking the address of b, you get char *. operator<< interprets that as a C string, and tries to print a character sequence instead of its address.
try cout << "address of char :" << (void *) &b << endl instead.
[EDIT] Like Tomek commented, a more proper cast to use in this case is static_cast, which is a safer alternative. Here is a version that uses it instead of the C-style cast:
cout << "address of char :" << static_cast<void *>(&b) << endl;
There are 2 questions:
Why it does not print the address for the char:
Printing pointers will print the address for the int*and the string* but will not print the contents for char* as there is a special overload in operator<<. If you want the address then use: static_cast<const void *>(&c);
Why the address difference between the int and the string is 8
On your platform sizeof(int) is 4 and sizeof(char) is 1 so you really should ask why 8 not 5. The reason is that string is aligned on a 4-byte boundary. Machines work with words rather than bytes, and work faster if words are not therefore "split" a few bytes here and a few bytes there. This is called alignment
Your system probably aligns to 4-byte boundaries. If you had a 64-bit system with 64-bit integers the difference would be 16.
(Note: 64-bit system generally refers to the size of a pointer, not an int. So a 64-bit system with a 4-byte int would still have a difference of 8 as 4+1 = 5 but rounds up to 8. If sizeof(int) is 8 then 8+1 = 9 but this rounds up to 16)
When you stream the address of a char to an ostream, it interprets that as being the address of the first character of an ASCIIZ "C-style" string, and tries to print the presumed string. You don't have a NUL terminator, so the output will keep trying to read from memory until it happens to find one or the OS shuts it down for trying to read from an invalid address. All the garbage it scans over will be sent to your output.
You can probably get it to display the address you want by casting it, as in (void*)&b.
Re the offsets into the structure: you observed the string is placed at offset 8. This is probably because you have 32-bit ints, then an 8-bit char, then the compiler chooses to insert 3 more 8-bit chars so that the string object will be aligned at a 32-bit word boundary. Many CPUs/memory-architectures need pointers, ints etc. to be on word-size boundaries to perform efficient operations on them, and would otherwise have to do many more operations to read and combine multiple values from memory before being able to use the values in an operation. Depending on your system, it may be that every class object needs to start on a word boundary, or it may be that std::string in particular starts with a size_t, pointer or other type that requires such alignment.
Because when you pass a char* to std::ostream it will print the C-style (ie: char array, char*) string it points to.
Remember that "hello" is a char*.
The address of char is being treated as a nul-terminated string and is displaying the contents of that address, which is probably undefined, but in this case an empty string. If you cast the pointers to void *, you will get the results you desire.
The difference between something2 and something being 8 is due to aligned and ability of the compiler to decide for itself where in the stack the variables are declared.
For the second issue - the compiler by default will pad structure members. The default pad is to the sizeof(int), 4 bytes (on most architectures). This is why an int followed by a char will take 8 bytes in the structure, so the string member is at offset 8.
To disable padding, use #pragma pack(x), where x is the pad size in bytes.
Your syntax should be
cout << (void*) &b
hrnt is right about the reason for the blank: &b has type char*, and so gets printed as a string until the first zero byte. Presumably b is 0. If you set b to, say, 'A', then you should expect the printout to be a string starting with 'A' and continuing with garbage until the next zero byte. Use static_cast<void*>(&b) to print it as a an address.
For your second question, &c - &i is 8, because the size of an int is 4, the char is 1, and the string starts at the next 8-byte boundary (you are probably on a 64-bit system). Each type has a particular alignment, and C++ aligns the fields in the struct according to it, adding padding appropriately. (The rule of thumb is that a primitive field of size N is aligned to a multiple of N.) In particular you can add 3 more char fields after b without affecting the address &c.
class Address {
int i ;
char b;
string c;
public:
void showMap ( void ) ;
};
void Address :: showMap ( void ) {
cout << "address of int :" << &i << endl ;
cout << "address of char :" << &b << endl ;
cout << "address of string :" << &c << endl ;
}
The output is:
address of int : something
address of char : // nothing, blank area, that is nothing displayed
address of string : something
Why?
Another interesting thing: if int, char, string is in public, then the output is
... int : something
... char :
... string : something_2
something_2 - something is always equal to 8. Why? (not 9)
When you are taking the address of b, you get char *. operator<< interprets that as a C string, and tries to print a character sequence instead of its address.
try cout << "address of char :" << (void *) &b << endl instead.
[EDIT] Like Tomek commented, a more proper cast to use in this case is static_cast, which is a safer alternative. Here is a version that uses it instead of the C-style cast:
cout << "address of char :" << static_cast<void *>(&b) << endl;
There are 2 questions:
Why it does not print the address for the char:
Printing pointers will print the address for the int*and the string* but will not print the contents for char* as there is a special overload in operator<<. If you want the address then use: static_cast<const void *>(&c);
Why the address difference between the int and the string is 8
On your platform sizeof(int) is 4 and sizeof(char) is 1 so you really should ask why 8 not 5. The reason is that string is aligned on a 4-byte boundary. Machines work with words rather than bytes, and work faster if words are not therefore "split" a few bytes here and a few bytes there. This is called alignment
Your system probably aligns to 4-byte boundaries. If you had a 64-bit system with 64-bit integers the difference would be 16.
(Note: 64-bit system generally refers to the size of a pointer, not an int. So a 64-bit system with a 4-byte int would still have a difference of 8 as 4+1 = 5 but rounds up to 8. If sizeof(int) is 8 then 8+1 = 9 but this rounds up to 16)
When you stream the address of a char to an ostream, it interprets that as being the address of the first character of an ASCIIZ "C-style" string, and tries to print the presumed string. You don't have a NUL terminator, so the output will keep trying to read from memory until it happens to find one or the OS shuts it down for trying to read from an invalid address. All the garbage it scans over will be sent to your output.
You can probably get it to display the address you want by casting it, as in (void*)&b.
Re the offsets into the structure: you observed the string is placed at offset 8. This is probably because you have 32-bit ints, then an 8-bit char, then the compiler chooses to insert 3 more 8-bit chars so that the string object will be aligned at a 32-bit word boundary. Many CPUs/memory-architectures need pointers, ints etc. to be on word-size boundaries to perform efficient operations on them, and would otherwise have to do many more operations to read and combine multiple values from memory before being able to use the values in an operation. Depending on your system, it may be that every class object needs to start on a word boundary, or it may be that std::string in particular starts with a size_t, pointer or other type that requires such alignment.
Because when you pass a char* to std::ostream it will print the C-style (ie: char array, char*) string it points to.
Remember that "hello" is a char*.
The address of char is being treated as a nul-terminated string and is displaying the contents of that address, which is probably undefined, but in this case an empty string. If you cast the pointers to void *, you will get the results you desire.
The difference between something2 and something being 8 is due to aligned and ability of the compiler to decide for itself where in the stack the variables are declared.
For the second issue - the compiler by default will pad structure members. The default pad is to the sizeof(int), 4 bytes (on most architectures). This is why an int followed by a char will take 8 bytes in the structure, so the string member is at offset 8.
To disable padding, use #pragma pack(x), where x is the pad size in bytes.
Your syntax should be
cout << (void*) &b
hrnt is right about the reason for the blank: &b has type char*, and so gets printed as a string until the first zero byte. Presumably b is 0. If you set b to, say, 'A', then you should expect the printout to be a string starting with 'A' and continuing with garbage until the next zero byte. Use static_cast<void*>(&b) to print it as a an address.
For your second question, &c - &i is 8, because the size of an int is 4, the char is 1, and the string starts at the next 8-byte boundary (you are probably on a 64-bit system). Each type has a particular alignment, and C++ aligns the fields in the struct according to it, adding padding appropriately. (The rule of thumb is that a primitive field of size N is aligned to a multiple of N.) In particular you can add 3 more char fields after b without affecting the address &c.
class Address {
int i ;
char b;
string c;
public:
void showMap ( void ) ;
};
void Address :: showMap ( void ) {
cout << "address of int :" << &i << endl ;
cout << "address of char :" << &b << endl ;
cout << "address of string :" << &c << endl ;
}
The output is:
address of int : something
address of char : // nothing, blank area, that is nothing displayed
address of string : something
Why?
Another interesting thing: if int, char, string is in public, then the output is
... int : something
... char :
... string : something_2
something_2 - something is always equal to 8. Why? (not 9)
When you are taking the address of b, you get char *. operator<< interprets that as a C string, and tries to print a character sequence instead of its address.
try cout << "address of char :" << (void *) &b << endl instead.
[EDIT] Like Tomek commented, a more proper cast to use in this case is static_cast, which is a safer alternative. Here is a version that uses it instead of the C-style cast:
cout << "address of char :" << static_cast<void *>(&b) << endl;
There are 2 questions:
Why it does not print the address for the char:
Printing pointers will print the address for the int*and the string* but will not print the contents for char* as there is a special overload in operator<<. If you want the address then use: static_cast<const void *>(&c);
Why the address difference between the int and the string is 8
On your platform sizeof(int) is 4 and sizeof(char) is 1 so you really should ask why 8 not 5. The reason is that string is aligned on a 4-byte boundary. Machines work with words rather than bytes, and work faster if words are not therefore "split" a few bytes here and a few bytes there. This is called alignment
Your system probably aligns to 4-byte boundaries. If you had a 64-bit system with 64-bit integers the difference would be 16.
(Note: 64-bit system generally refers to the size of a pointer, not an int. So a 64-bit system with a 4-byte int would still have a difference of 8 as 4+1 = 5 but rounds up to 8. If sizeof(int) is 8 then 8+1 = 9 but this rounds up to 16)
When you stream the address of a char to an ostream, it interprets that as being the address of the first character of an ASCIIZ "C-style" string, and tries to print the presumed string. You don't have a NUL terminator, so the output will keep trying to read from memory until it happens to find one or the OS shuts it down for trying to read from an invalid address. All the garbage it scans over will be sent to your output.
You can probably get it to display the address you want by casting it, as in (void*)&b.
Re the offsets into the structure: you observed the string is placed at offset 8. This is probably because you have 32-bit ints, then an 8-bit char, then the compiler chooses to insert 3 more 8-bit chars so that the string object will be aligned at a 32-bit word boundary. Many CPUs/memory-architectures need pointers, ints etc. to be on word-size boundaries to perform efficient operations on them, and would otherwise have to do many more operations to read and combine multiple values from memory before being able to use the values in an operation. Depending on your system, it may be that every class object needs to start on a word boundary, or it may be that std::string in particular starts with a size_t, pointer or other type that requires such alignment.
Because when you pass a char* to std::ostream it will print the C-style (ie: char array, char*) string it points to.
Remember that "hello" is a char*.
The address of char is being treated as a nul-terminated string and is displaying the contents of that address, which is probably undefined, but in this case an empty string. If you cast the pointers to void *, you will get the results you desire.
The difference between something2 and something being 8 is due to aligned and ability of the compiler to decide for itself where in the stack the variables are declared.
For the second issue - the compiler by default will pad structure members. The default pad is to the sizeof(int), 4 bytes (on most architectures). This is why an int followed by a char will take 8 bytes in the structure, so the string member is at offset 8.
To disable padding, use #pragma pack(x), where x is the pad size in bytes.
Your syntax should be
cout << (void*) &b
hrnt is right about the reason for the blank: &b has type char*, and so gets printed as a string until the first zero byte. Presumably b is 0. If you set b to, say, 'A', then you should expect the printout to be a string starting with 'A' and continuing with garbage until the next zero byte. Use static_cast<void*>(&b) to print it as a an address.
For your second question, &c - &i is 8, because the size of an int is 4, the char is 1, and the string starts at the next 8-byte boundary (you are probably on a 64-bit system). Each type has a particular alignment, and C++ aligns the fields in the struct according to it, adding padding appropriately. (The rule of thumb is that a primitive field of size N is aligned to a multiple of N.) In particular you can add 3 more char fields after b without affecting the address &c.
#include <cstdlib>
#include <iostream>
int main(int argc, char *argv[])
{
cout << "size of String " << sizeof( string );
system("PAUSE");
return EXIT_SUCCESS;
}
Output:
size of String = 4
Does that mean that, since sizeof(char) = 1 Byte (0 to 255), string can only hold 4 characters?
It isn't clear from your example what 'string' is. If you have:
#include <string>
using namespace std;
then string is std::string, and sizeof(std::string) gives you the size of the class instance and its data members, not the length of the string. To get that, use:
string s;
cout << s.size();
When string is defined as:
char *string;
sizeof(string) tells you the size of the pointer. 4 bytes (You're on a 32-bit machine.) You've allocated no memory yet to hold text. You want a 10-char string? string = malloc(10); Now string points to a 10-byte buffer you can put characters in.
sizeof(*string) will be 1. The size of what string is pointing to, a char.
If you instead did
char string[10];
sizeof(string) would be 10. It's a 10-char array.
sizeof(*string) would be 1 still.
It'd be worth looking up and understanding the __countof macro.
Update: oh, yeah, NOW include the headers :) 'string' is a class whose instances take up 4 bytes, that's all that means. Those 4 bytes could point to something far more useful, such as a memory area holding more than 4 characters.
You can do things like:
string s = "12345";
cout << "length of String " << s.length();
sizeof(char) is always 1 byte. A byte which we think is 8-bits need not be the case. There are architectures where a BYTE is 32-bits, 24-bits and so on. The sizeof applied to any other type is in multiples of sizeof(char) which is by definition 1.
The next important thing to note is that C++ has three character types: plain char, signed char and unsigned char. A plain char is either signed or unsigned. So it is wrong to assume that char can have only values from 0 to 255. This is true only when a char is 8-bits, and plain char is unsigned.
Having said, that assuming that 'string' is 'std::namespace', sizeof(string) == 4 means that the sizeof the 'std::string' class is 4 bytes. It occupies 4 times the number of bytes that a 'char' on that machine takes. Note that signed T, unsigned T always have the same size. It does not mean that the actual buffer of characters (which is called string in common parlance) is only 4 bytes. Inside the 'std::string' class, there is a non static member pointer which is allocated dynamically to hold the input buffer. This can have as many elements as the system allows (C++ places no restriction on this length). But since the 'std::string' class only holds the pointer to this potentially infite length buffer, the sizeof(std::string) always remains the same as sizeof pointer on the given architecture which on your system is 4.
I know a lot of people had answered your question, but here are some points:
It's not the size of the string or the capacity of the string, this value represents the structural size of the class string, which you can see by its implementation (and it can change from implementation to implementation) that is a simple pointer;
As the sizeof(string) is the size of the class structure, you'll get the size of the only internal pointer, that in your case is 4 bytes (because you are in a 32-bit machine, this can change from platform to platform too);
This pointer inside the string class, points to a memory buffer where the class will hold the real string data, this memory buffer is reallocated as needed, it can increase/decrease as you append/delete/create more string text;
If you want to get the real size of the string, you need to call the size() method from the class which will check the memory buffer string size (which isn't the same as the memory buffer size).
I think your problem is your conception of sizeof, see more information here and here is some explanation on how it works.
Not at all. It means that the class's structure is that, it doesn't include the dynamic memory it can control. std::string will expand dynamically to meet any required size.
s.max_size() // will give the true maximum size
s.capacity() // will tell you how much it can hold before resizing again
s.size() // tells you how much it currently holds
The 4 you get from sizeof is likely a pointer of some kind to the larger structure. Although some optimizations on some platforms will use it as the actual string data until it grows larger than can fit.
No, it means that the sizeof the class string is 4.
It does not mean that a string can be contained in 4 bytes of memory. Not at all. But you have to difference between dynamic memory, used to contain the size characters a string can be made of, and the memory occupied by the address of the first of those characters
Try to see it like this:
contents --------> |h|e|l|l|o| |w|o|r|ld|\0|
sizeof 4 refers to the memory occupied by contents. What it contents? Just a pointer to (the address of ) the first character in the char array.
How many characters does a string can contain ? Ideally, a character per byte available in memory.
How many characters does a string actually have? Well, theres a member function called size() that will tell you just that
size_type size() const
See more on the SGI page !
A string object contains a pointer to a buffer on the heap that contains the actual string data. (It can also contain other implementation-specific meta-information, but yours apparently doesn't.) So you're getting the size of that pointer, not the size of the array it points to.
you can also use strings and can find out its length by string.length() function. look at the below code:
// Finding length of a string in C++
#include<iostream>
#include<string>
using namespace std;
int count(string);
int main()
{
string str;
cout << "Enter a string: ";
getline(cin,str);
cout << "\nString: " << str << endl;
cout << count(str) << endl;
return 0;
}
int count(string s){
if(s == "")
return 0;
if(s.length() == 1)
return 1;
else
return (s.length());
}
you can get the details from :
http://www.programmingtunes.com/finding-length-of-a-string-in-c/
size() of string gives the number of elements in the string whereas sizeof() function on a string gives three extra bits. strlen() of a character array gives the number of elements + 1 (because of null char delimiter) and keep in mind size of char is 1 byte. sizeof() on a char array gives the size assigned to the array
string str="hello";
char arr[x]="hello";
cout<<str.size()<<endl<<sizeof(str)<<endl;
cout<<strlen(arr)<<endl<<sizeof(arr)<<endl;
output is 5 8 5 x