I am writing my own append function to append a dynamic character array of string array2 at the end of another dynamic character array of string array1, using a static char buffer[50]. But the compiler generates the following errors: [Error] incompatible types in assignment of 'char' to 'char[50]'. I have tried to figure out the problem, but I don't seem to find the solution. Your help will be very appreciated. I am using Dev-C++. The code is bellow.
#include <iostream>
using namespace std;
char *Appendstring(char *a, char *b) // will append b to the end of a
{
static char buffer[50];
char *p=buffer=*a++; //[Error] incompatible types in assignment of 'char' to 'char[50]'
//[Error] invalid conversion from 'char*' to 'char'[-fpermissive]
p--;
while(*p++=b++);
p--; //append
while(*p++=*c++);
return buffer;
}
int main ()
{
string str="Displaying: ";
string add=" Summer is coming";
Appendstring(str, add);
return 0;
}
There are multiple errors in your append function, the biggest ones are using an array as a pointer and using a static buffer to merge strings. With a static buffer in place, all your merged strings will be in the same space, so merging two strings and then merging the other two would overwrite the results of the first merge!
You can change your function as follows:
char *Appendstring(const char *a, const char *b) // will append b to the end of a
{
char *buffer = new char[strlen(a)+strlen(b)+1];
char *p=buffer;
while(*p++=*a++); // Copy a into buffer
while(*p++=*b++); // Copy b into buffer right after a
*p=0; // Null-terminate the string
return buffer;
}
Of course the caller is responsible for freeing the results of Appendstring now.
You cannot assign into an array, which is what you do in buffer=*a++. what you meant is probably
static char buffer[50];
char *p=buffer;
*p=*a++;
In addition, here
p--;
while(*p++=*b++);
you are trying to derefence a pointer one element before the beginning of an array - which leads to undefined behaviour.
Moreover, nowhere do you check for the strings' length, so it can easily be more the 49 together and your code will be both incorrect and insecure (easy victim for buffer overflow attacks).
One last problem is that your code is non reentrant in any way, due to the use of static array. you can simply use simple array, if you don't want to adjust it to the strings' length, or allocate it dynamically, as was suggested here.
The best solution of course is to use std::string and forget all these problems.
Related
First of all I am beginner in C++. I was trying to learn about type casting in C++ with strings and character pointer. Is it possible to point a string with a character pointer?
int main() {
string data="LetsTry";
cout<<(&data)<<"\n";
cout<<data<<"\n"<<"size "<<sizeof(data)<<"\n";
//char *ptr = static_cast<char*>(data);
//char *ptr=(char*)data;
char *ptr = reinterpret_cast<char*>(&data);
cout<<(ptr)<<"\n";
cout<<*ptr;
}
The above code yields outcome as below:
0x7ffea4a06150
LetsTry
size 32
`a���
`
I understand as ptr should output the address 0x7ffea4a06150
Historically, in C language strings were just a memory areas filled with characters. Consequently, when a string was passed to a function, it was passed as a pointer to its very first character, of type char *, for mutable strings, or char const *, if the function had no intent to modify string's contents. Such strings were delimited with a zero-character ((char)0 a.k.a. '\0') at the end, so for a string of length 3 you had to allocate at least four bytes of memory (three characters of the string itself plus the zero terminator); and if you only had a pointer to a string's start, to know the size of the string you'd have to iterate it to find how far is the zero-char (the standard function strlen did it). Some standard functions accepted en extra parameter for a string size if you knew it in advance (those starting with strn or, more primitive and effective, those starting with mem), others did not. To concatenate two strings you first had to allocate a sufficient buffer to contain the result etc.
The standard functions that process char pointers can still be found in STL, under the <cstring> header: https://en.cppreference.com/w/cpp/header/cstring, and std::string has synonymous methods c_str() and data() that return char pointers to its contents, should you need it.
When you write a program in C++, its main function has the header of int main(int argc, char *argv[]), where argv is the array of char pointers that contains any command-line arguments your program was run with.
Ineffective as it is, this scheme could still be regarded as an advantage over strings of limited capacity or plain fixed-size character arrays, for instance in mid-nineties, when Borland introduced the PChar type in Turbo Pascal and added a unit that exported Pascal implementations of functions from C's string.h.
std::string and const char* are different types, reinterpret_cast<char*>(&data) means reinterpret the bits located at &data as const char*, which is not we want in this case.
so assuming we have type A and type B:
A a;
B b;
the following are conversion:
a = (A)b; //c sytle
// and
a = A(b);
// and
a = static_cast<A>(b); //c++ style
the following are bit reinterpretation:
a = *(A*)&b; //c style
// and
a = *reinterpret_cast<A*>(&b); //c++ style
finally, this should works:
int main() {
string data = "LetsTry";
const char *ptr = data.c_str();
cout<< ptr << "\n";
}
bit reinterpretation is sometimes used, like when doing bit manipulation of a floating point number, but there are some rules to follow like this one What is the strict aliasing rule?
also note that cout << ptr << "\n"; is a specially case because feeds a pointer to std::cout usually output the address that pointer points to, but std::cout treats char* specially so that it output the content of that char array instead
In C++, string is class and what you doing is creating a string object. So, to use are char * you need to convert it using c_str()
You can refer below code:
std::string data = "LetsTry";
// declaring character array
char * cstr = new char [data.length()+1];
// copying the contents of the
// string to char array
std::strcpy (cstr, data.c_str());
Now, you can get use char * to point your data.
I'm working with a piece of C++ code that reads lines from a txt file and then assign each line to to an array called lines. Then it calls a function that converts each element in the lines array to a char array and then return the resulted char array. This step is where I stuck. How could I return a char array from the function toChar and assign the returned array to another array so I can use it as I need? (the rest of the code should use each returned char array to write it in a pipe, this not important right now but just to clarify why I need to learn to return an array from a function)
Here is the code I'm using:
#include <fstream>
#include <iostream>
#include <string.h>
#include <unistd.h>
#include <cstdlib>
using namespace std;
char * toChar(string line);
int main()
{
string lines[0] = "line1";
char* a = toChar(lines[0]);
return 0;
}
char * toChar(string line)
{
char a[1024];
strcpy(a, line.c_str());
return a;
}
Please note that in this code I'm trying to shrink the code so I'm assigning a simple string value to the array
when I try to compile this code, the error below appears:
warning: address of local variable 'a' returned
any help or suggestion is greatly appreciated..
First use a const & in passing the string to the function to avoid the unnecessary copying, and be able to use temporaries, e.g. toChar("I am temp");
The are the following alternatives:
(1) Return std::string
std::string toChar(std::string const & line)
{
char a[1024];
strcpy(a, line.c_str());
return a;
}
Of course it is assumed that line is smaller than 1024 chars with the null termination symbol
(2) Return an allocated array the
char * toChar(std::string const & line)
{
char * a = new char[1024];
strcpy(a, line.c_str());
return a;
}
but you will have to actually manage it and delete it later.
(3) Allocate the array and pass it to the function
void toChar(string const & line, char a[])
{
strcpy(a, line.c_str());
}
I imagine that you actually want to extract a C-string from an std::string , or some part of it. The proper way to do it is (3).
The warning is correct and practically is an error. You're declaring the char array locally so it will be deleted after going out of scope of that function and its address will be no more valid.
Since you're using c++ avoid char array and use a std::string.
If you want to access to the internal storage of your string you can call line.c_str() which returns a const char *.
Hence you won't need you function toChar that creates a local array that will go out of scope at the end of your function.
You could simply do :
int main()
{
string lines[0] = "line1";
const char* a = lines[0].c_str();
return 0;
}
But I advise you to keep manipulating std::string as they will handle string better than a simple char *.
And if you want a copy, just do it : std::string mycopy = lines[0].
Just trying to assign chars to the char array and it says string in not null terminated?
I want to be able to change the teams around in the array like a scoreboard.
#include <string.h>
#include <iostream>
int main(int argc, char* argv[])
{
char Team1[7] = "Grubs";
char Team2[7] = "Giants";
char Team3[7] = "Bulls";
char Team4[7] = "Snakes";
char Team5[7] = "Echos";
char TeamList[5][7];
strcpy_s(TeamList[0], Team1);
strcat_s(TeamList[1], Team2);
strcat_s(TeamList[2], Team3);
strcat_s(TeamList[3], Team4);
strcat_s(TeamList[4], Team5);
TeamList[5][7]= '\0';
system("pause");
return 0;
}
strcat() (which is a "less-safe" version of strcat_s()) requires both strings to be null-terminated. That's because strcat() appends its second parameter (source) where first parameter (dest) ends. It replaces null-terminator of dest with first character of source, appends rest of source and then
a null-character is included at the end of the new string formed by
the concatenation of both
I would simply change
strcpy_s(TeamList[0], Team1);
strcat_s(TeamList[1], Team2);
strcat_s(TeamList[2], Team3);
strcat_s(TeamList[3], Team4);
strcat_s(TeamList[4], Team5);
to
strcpy_s(TeamList[0], Team1);
strcpy_s(TeamList[1], Team2);
strcpy_s(TeamList[2], Team3);
strcpy_s(TeamList[3], Team4);
strcpy_s(TeamList[4], Team5);
strcpy_s() does not have any requirements regarding contents of destination - only its capacity matters.
If you want to stick with strcat_s(), do this:
char TeamList[5][7];
memset(TeamList, 0, sizeof(char) * 5 * 7);
Then, this line:
TeamList[5][7]= '\0';
is not required, It is incorrect anyway, because for N-element array valid indexes are [0; N-1].
EDIT
Since in your case swapping comes into play, I would suggest you totally different approach.
First of all:
#include <string>
Then, initialize teams this way:
std::string TeamList[] =
{
"Grubs",
"Giants",
"Bulls",
"Snakes",
"Echos"
};
Now, TeamList is an array containing 5 elements and each of these elements is an object of type std::string, containing name of a particular team.
Now, if you want to swap, let's say, teams 1 and 3:
std::swap(TeamList[1], TeamList[3]);
std::swap() is a standard C++ function extensively used in standard library implementation. It is overloaded for many standard types, including std::string. This solution has one, critical benefit: if string's content is held on the heap, swapping two strings is as simple as swapping pointers (and some length/capacity variables).
Oh, and one more thing: if you are not familiar with std::string and you would need to get pointer to a buffer containing string's data, you can do it this way:
const char* team_1_raw_name = TeamList[0].c_str();
See this page for more info about std::string
strcat requires that there already be a null-terminated string in the destination to concatenate the source string onto; you're calling it with uninitialised values in the destination.
It looks like you want strcpy in every case, not just the first.
Also, remove the bogus TeamList[5][7]= '\0';. Even if you fix it to write inside the array bounds, each string has already been terminated by strcpy so there's no need to try to do that yourself.
Then stop messing around with low-level arrays and pointers. std::vector<std::string> would be much friendlier.
I am trying to make a function like strcpy in C++. I cannot use built-in string.h functions because of restriction by our instructor. I have made the following function:
int strlen (char* string)
{
int len = 0;
while (string [len] != (char)0) len ++;
return len;
}
char* strcpy (char* *string1, char* string2)
{
for (int i = 0; i<strlen (string2); i++) *string1[i] = string2[i];
return *string1;
}
main()
{
char* i = "Farid";
strcpy (&i, "ABC ");
cout<<i;
}
But I am unable to set *string1 [i] value. When I try to do so an error appears on screen 'Program has encountered a problem and need to close'.
What should I do to resolve this problem?
Your strcpy function is wrong. When you write *string1[i] you are actually modifying the first character of the i-th element of an imaginary array of strings. That memory location does not exist and your program segfaults.
Do this instead:
char* strcpy (char* string1, char* string2)
{
for (int i = 0; i<strlen (string2); i++) string1[i] = string2[i];
return string1;
}
If you pass a char* the characters are already modifiable. Note It is responsibility of the caller to allocate the memory to hold the copy. And the declaration:
char* i = "Farid";
is not a valid allocation, because the i pointer will likely point to read-only memory. Do instead:
char i[100] = "Farid";
Now i holds 100 chars of local memory, plenty of room for your copy:
strcpy(i, "ABC ");
If you wanted this function to allocate memory, then you should create another one, say strdup():
char* strdup (char* string)
{
size_t len = strlen(string);
char *n = malloc(len);
if (!n)
return 0;
strcpy(n, string);
return n;
}
Now, with this function the caller has the responsibility to free the memory:
char *i = strdup("ABC ");
//use i
free(i);
Because this error in the declaration of strcpy: "char* *string1"
I don't think you meant string1 to be a pointer to a pointer to char.
Removing one of the * should word
The code has several issues:
You can't assign a string literal to char* because the string literal has type char const[N] (for a suitable value of N) which converts to char const* but not to char*. In C++03 it was possible to convert to char* for backward compatibility but this rule is now gone. That is, your i needs to be declared char const*. As implemented above, your code tries to write read-only memory which will have undesirable effects.
The declaration of std::strcpy() takes a char* and a char const*: for the first pointer you need to provide sufficient space to hold a string of the second argument. Since this is error-prone it is a bad idea to use strcpy() in the first place! Instead, you want to replicate std::strncpy() which takes as third argument the length of the first buffer (actually, I'm never sure if std::strncpy() guarantees zero termination or not; you definitely also want to guarantee zero termination).
It is a bad idea to use strlen() in the loop condition as the function needs to be evaluated for each iteration of the loop, effectively changing the complexity of strlen() from linear (O(N)) to quadratic (O(N2)). Quadratic complexity is very bad. Copying a string of 1000 characters takes 1000000 operations. If you want to try out the effect, copy a string with 1000000 characters using a linear and a quadratic algorithm.
Your strcpy() doesn't add a null-terminator.
In C++ (and in C since ~1990) the implicit int rule doesn't apply. That is, you really need to write int in front of main().
OK, a couple of things:
you are missing the return type for the main function
declaration. Not really allowed under the standard. Some compilers will still allow it, but others will fail on the compile.
the way you have your for loop structured in
strcpy you are calling your strlen function each time through
the loop, and it is having to re-count the characters in the source
string. Not a big deal with a string like "ABC " but as strings get
longer.... Better to save the value of the result into a variable and use that in the for loop
Because of the way that you are declaring i in
`main' you are pointing to read-only storage, and will be causing an
access violation
Look at the other answers here for how to rebuild your code.
Pointer use in C and C++ is a perennial issue. I'd like to suggest the following tutorial from Paul DiLorenzo, "Learning C++ Pointers for REAL dummies.".
(This is not to imply that you are a "dummy," it's just a reference to the ",insert subject here> for Dummies" lines of books. I would not be surprised that the insertion of "REAL" is to forestall lawsuits over trademarked titles)
It is an excellent tutorial.
Hope it helps.
I have followed the code example here
toupper c++ example
And implemented it in my own code as follows
void CharString::MakeUpper()
{
char* str[strlen(m_pString)];
int i=0;
str[strlen(m_pString)]=m_pString;
char* c;
while (str[i])
{
c=str[i];
putchar (toupper(c));
i++;
}
}
But this gives me the following compiler error
CharString.cpp: In member function 'void CharString::MakeUpper()':
CharString.cpp:276: error: invalid conversion from 'char*' to 'int'
CharString.cpp:276: error: initializing argument 1of 'int toupper(int)'
CharString.cpp: In member function 'void CharString::MakeLower()':
This is line 276
putchar (toupper(c));
I understand that toupper is looking for int as a parameter and returns an int also, is that the problem? If so how does the example work?
Also,
char* str[strlen(m_pString)];
int i=0;
str[strlen(m_pString)]=m_pString;
is not valid C++ - arrays must be dimensioned using compile time constants - this is a C99 feature. And I really don't think the code would do what you want it to, even if it were legal, as you seem to be accessing one past the end of the array. It would be handy if you posted the complete class definition.
I don't think your code does what you want it to do and in fact if it compiled it would explode.
char* str[strlen(m_pString)]; // you've made an array of X C strings where
// X is the length of your original string.
int i=0;
str[strlen(m_pString)]=m_pString; // You've attempted to assign the C string in your array
// at location X to point at you m_pString. X is the
// same X as before and so is 1 past the end of the array
// This is a buffer overrun.
I think what you actually wanted to do was to copy the content of m_pString into str. You'd do that like so:
char * str = new char[strlen(m_pString)];
memcpy(str, m_pString); // I may have the operands reversed, see the docs.
The easier way to do this though is to stop using C strings and to use C++ strings:
std::string str = m_pString;
There are more issues, but this should get you steer you more toward the right direction.
You need to feed toupper() an int (or a char) instead of a char *, which is how you've declared c.
try:
char c;
Also,
char* str[strlen(m_pString)];
is an an array of pointers to characters, not just a single string.
This line:
str[strlen(m_pString)]=m_pString;
is an assignment to a bad pointer then, since there was no allocation.
I'm going to go with the assumption that m_pString is a C style string (char *). You're doing way more fiddling than you need to be doing.
void CharString::MakeUpper()
{
char* str = m_pString; // Since you're not modifying the string, there's no need to make a local copy, just get a pointer to the existing string.
while (*str) // You can use the string pointer as an iterator over the individual chars
{
putchar (toupper(*str)); // Dereference the pointer to get each char.
str++; // Move to the next char (you can merge this into the previous line if so desired, but there's no need.
}
}
In the example you cite, the reason it works is because of how the variables are declared.
int main ()
{
int i=0;
char str[]="Test String.\n"; // This is a compile time string literal, so it's ok to initialize the array with it. Also, it's an array of `char`s not `char*`s.
char c; // Note that this is also a `char`, not a `char *`
while (str[i])
{
c=str[i];
putchar (toupper(c));
i++;
}
return 0;
}
Because of the error-prone ways of using C strings, your best bet is std::string:
void CharString::MakeUpper()
{
string str(m_pString);
transform(str.begin(), str.end(), ostream_iterator<char>(cout), &toupper);
}
There is not a built-in conversion from char * to int, which is why the error occurs. Since you're trying to capitalize a character, you need to dereference the pointer.
putchar(toupper(*c));