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#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
I am working on a coding assignment for my class and I ran into a problem!
I have this constructor here, for a String object:
String::String(char str[]) {
size = (sizeof(str)/sizeof(str[0]));
data = new char[size];
for (int i = 0; i < size; ++i) {
data[i] = str[i];
}
}
Here is part of the main I was provided:
char test[11] = "Hello world";
String two(test);
cout << "The length of String two is: " <<
two.length() << endl;
cout << "The value of String two is: ";
two.print();
So when I run this, I would get 8 for the size (should be 11). However, after some research, I figured out it is because the sizeof(str) is returning the byte size of a pointer, rather than the entire array.
So is there any way to get the size of the whole array with what I have? I am not supposed to manipulate the provided main, therefore I cannot add an int size to the parameters, which would be the obvious solution.
I've been stuck on this one for a bit, thanks for any help and suggestions,
Array decays to pointer when passed to a function.
You have to either pass the length to the function, pass a STL container e.g. std::vector or use strlen() inside function. (Note that strlen() need a terminating null-character to work properly and you have to add that to your array)
You can not get size of array at runtime in C. At runtime, array is just the address. The size is simply not stored anywhere. In source code, at compile time, in a place where compiler knows the size, you can use sizeof operator, but that gets essentially converted to a constant numeric literal, ie. same as writing the right number there yourself (VLAs are a bit more complex case, and of course using sizeof can create portable code unlike hard-coded number).
To make matters worse (for understanding C), when you have a function parameter that looks like an array, it really is a pointer. Even if you give it static size in the parameter list, sizeof still it gives you size of pointer, for example. Only non-parameter variables can actually be arrays, with sizeof working as expected.
You have to pass the size somehow (usually as extra parameter) or have some other way of telling where the data ends (such as strings' '\0' at the end).
Use a vector instead of char array. You can get size by calling size() method of vector container. If you want to use a char array, then it is a common practice in c programming to pass size as second parameter in the function.
You will only get size of array using sizeof() function on the function stack in which the array is defined and if the array size is known in compile time.
I was working with a program that uses a function to set a new value in the registry, I used a const char * to get the value. However, the size of the value is only four bytes. I've tried to use std::string as a parameter instead, it didn't work.
I have a small example to show you what I'm talking about, and rather than solving my problem with the function I'd like to know the reason it does this.
#include <iostream>
void test(const char * input)
{
std::cout << input;
std::cout << "\n" << sizeof("THIS IS A TEST") << "\n" << sizeof(input) << "\n";
/* The code above prints out the size of an explicit string (THIS IS A TEST), which is 15. */
/* It then prints out the size of input, which is 4.*/
int sum = 0;
for(int i = 0; i < 15; i++) //Printed out each character, added the size of each to sum and printed it out.
//The result was 15.
{
sum += sizeof(input[i]);
std::cout << input[i];
}
std::cout << "\n" << sum;
}
int main(int argc, char * argv[])
{
test("THIS IS A TEST");
std::cin.get();
return 0;
}
Output:
THIS IS A TEST
15
4
THIS IS A TEST
15
What's the correct way to get string parameters? Do I have to loop through the whole array of characters and print each to a string (the value in the registry was only the first four bytes of the char)? Or can I use std::string as a parameter instead?
I wasn't sure if this was SO material, but I decided to post here as I consider this to be one of my best sources for programming related information.
sizeof(input) is the size of a const char* What you want is strlen(input) + 1
sizeof("THIS IS A TEST") is size of a const char[]. sizeof gives the size of the array when passed an array type which is why it is 15 .
For std::string use length()
sizeof gives a size based on the type you give it as a parameter. If you use the name of a variable, sizeof still only bases its result on the type of that variable. In the case of char *whatever, it's telling you the size of a pointer to char, not the size of the zero-terminated buffer it's point at. If you want the latter, you can use strlen instead. Note that strlen tells you the length of the content of the string, not including the terminating '\0'. As such, if (for example) you want to allocate space to duplicate a string, you need to add 1 to the result to tell you the total space occupied by the string.
Yes, as a rule in C++ you normally want to use std::string instead of pointers to char. In this case, you can use your_string.size() (or, equivalently, your_string.length()).
std::string is a C++ object, which cannot be passed to most APIs. Most API's take char* as you noticed, which is very different from a std::string. However, since this is a common need, std::string has a function for that: c_str.
std::string input;
const char* ptr = input.c_str(); //note, is const
In C++11, it is now also safe-ish to do this:
char* ptr = &input[0]; //nonconst
and you can alter the characters, but the size is fixed, and the pointer is invalidated if you call any mutating member of the std::string.
As for the code you posted, "THIS IS A TEST" has the type of const char[15], which has a size of 15 bytes. The char* input however, has a type char* (obviously), which has a size of 4 on your system. (Might be other sizes on other systems)
To find the size of a c-string pointed at by a char* pointer, you can call strlen(...) if it is NULL-terminated. It will return the number of characters before the first NULL character.
If the registry you speak of is the Windows registry, it may be an issue of Unicode vs. ASCII.
Modern Windows stores almost all strings as Unicode, which uses 2 bytes per character.
If you try to put a Unicode string into an std::string, it may be getting a 0 (null), which some implementations of string classes treat as "end of string."
You may try using a std::wstring (wide string) or vector< wchar_t > (wide character type). These can store strings of two-byte characters.
sizeof() is also not giving you the value you may think it is giving you. Your system probably runs 32-bit Windows -- that "4" value is the size of the pointer to the first character of that string.
If this doesn't help, please post the specific results that occur when you use std::string or std::wstring (more than saying that it doesn't work).
To put it simply, the size of a const char * != the size of a const char[] (if they are equal, it's by coincidence). The former is a pointer. A pointer, in the case of your system, is 4 bytes REGARDLESS of the datatype. It could be int, char, float, whatever. This is because a pointer is always a memory address, and is numeric. Print out the value of your pointer and you'll see it's actually 4 bytes. const char[] now, is the array itself and will return the length of the array when requested.
#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
I am trying to get the size of an array populated by stdin:
char *myArray;
cin >> myArray
cout << sizeof(myArray);
This returns 4 when I enter a string greater with a length greater than 4 e.g. "40905898"
Where am i going wrong?
sizeof operator statically evaluates to the size of the thing you are passing to it. A char* is a pointer which, depending on the machine architecture has a specific size (4 bytes on 32 bit systems and 8 bytes on 64 bit machines). To accomplish what you are trying to do, I suggest you use the string type which you can use by adding #include <string> along with using namespace std; to your source file.
string line;
cin >> line;
cout << line.length() << endl;
It's less error prone and easier to use.
By the way, the thing you've tried to do is really dangerous. In fact, when you use cin >> myArray, you should have already allocated some memory for myArray which you haven't done. This will cause memory corruption which might crash your program and possibly put it to buffer overrun attacks.
A simple array in C++ has no idea about its size. You can use sizeof only if the array is statically allocated and you are using sizeof on the array itself, not another pointer to it, for example this won't work as you might expect:
int x[5];
int *a = &x[0];
// a[i] is now the same as x[i] but:
cout << sizeof(x) << endl; // prints 20, assuming int is 32 bits long
cout << sizeof(a) << endl; // prints 4, assuming a pointer is 32 bits long
Note that the total size of the array is printed on the first line, not the element count. You can use sizeof(x)/sizeof(*x) to find out element count in static arrays. This thing is not possible for dynamically allocated arrays using new. In fact C++ arrays are very error prone and you should take extreme care when working with them and you'd better use vector and string in most cases instead.
sizeof(pointer) will always return 4. You want to use strlen().
Edit: IIRC, sizeof is evaluated at compile time, it only cares about the type, not the content.
This is because myArray is a pointer that occupies 4 bytes. If you want to get the length of your string, use strlen or something similar.
It's because you are using sizeof() on a pointer, which is 4 bytes on your 32-bit computer:
printf("Pointer size: %d\n", sizeof(void*));
If your array is a null-terminated string (the last element being a zero-byte, or '\0'), then you can use
strlen(myArray)
to get the number of elements (minus one). E.g.:
myArray = "Hello, world!";
printf("Number of characters: %d\n", strlen(myArray));
You could also use a statically allocated array, like this:
char array[128];
printf("sizeof(array) = %d\n", sizeof(array));
// prints 128
As others said, myArray is a pointer.
But why wouldn't you use std::string? You won't need to do the buffer allocation yourself, which you do wrongly in your example (pointer myArray points to nothing)
std::string myValue;
std::cin >> myValue;
std::cout << myValue.length();
If needed, you can get to a pointer representation of the string by using string::c_str().
As others said sizeof returns the size of the object passed to it, in the case of a pointer it's the size of the pointer. sizeof does not follow the pointer to see what size the object it points to is (how can it know it could point to one char or to an array there's not good way to know).
Additionally when you read from cin you have to allocate some space into which the data must be read, cin will not allocate space for you. You can allocate space either on the stack or the heap:
char stack_line[1024]; // This will overflow if more than 1024 chars are needed
char heap_line* = new char[1024]; // ditto
The thing to note here is that sizeof(stack_line) == 1024 while sizeof(heap_line) == 4 [on 32 bit machines] so you must be careful when using the sizeof operator.
In practice it's better to use an std::string which knows how to allocate the space itself.
You seem to have lots of problems here:
myArray is not initialised - where is the input going to live?
You usually use: cin >> myArray; (Note the direction of the chevrons and the semi-colon)
sizeof(myArray) will always return the same value (4 on your platform)
Try this version instead:
char* myArray= new char[50];
cin >> myArray;
cout << myArray;
cout << strlen(myArray);
Its not without its own problems (I should have deleted myArray), so you should try the answers here that use string myArray
well if you were going to use sizeof(myArray), you should've done sizeof(*myArray); because with pointers, no star is the address not the value.
Like this:
char *myArray;
cin >> *myArray;
cout << *myArray;