Coincise question: Need to be able to modify contents of char* in c++.
I have a function which is somewhat like this: char* buffer = (char*)FUNCTION
Now I need to modify 'buffer' e.g. do something like this buffer[1] = "h";
Amongst what I have tried is the following:
char *string = "Hello World";
char buffer[65]; //I still need to know the exact size
strcpy(buffer, string);
buffer[1] = "r";
I have tried with malloc also.
Unfortunately, the compiler always complain with the following: "cannot convert from const char[2] to char". This occurs on windows.
I do not have such a problem with g++ compiler.
I have had a looked at the following links but still I can't do anything.
http://www.developerweb.net/forum/archive/index.php/t-3517.html
About character pointers in C
Is it possible to modify a string of char in C?
Thanks
Since your question is tagged C++, I must question your intentions to be using raw char*s, it's not recommended unless you're very confident.
Secondly, a char* or char[] is perfectly valid to modify. The problem that you've got is that you made a const char* in "Hello World", then tried to make one character in it equal to a string. A character is not a string. "r" is a string, it's a const char*. So you're trying to equate a const char* to a char, which is an error. You really mean 'r', which is a char.
Better code would be:
std::string mystring((char*)FUNCTION); // assume that FUNCTION
// will free it's own memory.
if (index < mystring.size())
mystring[index] = 'r';
// mystring now cleans up it's own memory in all cases.
You can't change *char string = "Hello World" as string is "read-only" but you can use char string[] = .. which will compiler arrange into modifiable array:
char * strReadOnly = "read-only-data";
//strReadOnly [3] = '0'; //access violation
char stReadWrite [] = "read-write-data";
stReadWrite [3] = '0'; // Ok, but just make sure that indexes are in range
To copy string into another buffer that you can change, you can do following:
char * src = "read-only-data";
const int len = strlen(src);
char * dst = new char[len+1];
strcpy(dst, src);
dst[3] = '0'; // ok to change
delete [] dst; // don't forget to delete dst
Your problem is in quotes, you must use '' if you want char from compiler, "" is a string and have anditional null termination (one char) then you get error char[2] (char + null) to char.
Try *(buffer + x) = 'h';
x - is your position of char to change.
You have to understand that there is a difference between the following two lines:
char myArray[] = "Hello Mars!";
char* myPointer = "Hello World!";
The first one creates an array of chars, which can be manipulated. The whole array will be allocated on the stack.
The second one initializes a pointer to char with a string literal (a special construct from the C language allows this). This is to be considered a constant. You will get the address of it, but it is an array of const chars really. The implementation is as follows:
The string will be stored literally in your executable image in memory (as an identical copy of the file on hdd), and you will get the pointer to that string. Space will be made on the stack for the pointer myPointer. Consider the string literal like a literal number in your code (yes, the magic numbers I'm talking about). The compiler needs to store those characters somewhere, but it is not the kind of memory you will normally work with in your program. It is not on the heap, nor on the stack. See the answer to the following question for more info:
Global memory management in C++ in stack or heap?
edit: about pointers and arrays
It is not the char* that makes it immutable, but the string literal. Pointers and arrays are very much related. An array will very often behave like a pointer. Considering the code above this is completely legal:
myPointer = myArray;
myPointer was not const and you can change it to point to the address of myArray instead of the string literal. The address it pointed to was read-only. Now myPointer points to the adress on the stack where "Hello Mars!" is stored. You could now write:
myPointer[3] = 'Z';
If you want to get a torough understanding of this, read some tutorial like this one.
on constness
Const can be confusing especially with pointers.
const char* const myPointer2 = myPointer;
The first const makes the characters const, so you cannot use myPointer2 to change them, and the second const makes the pointer itself const, so you cannot let it point to something else like myPointer2 = myArray;
See this explanation for more info.
