Thread: Understang this malloc() related code

Please indicate to me if I have the general idea of this memory allocating code correct. Thanks

Code:

char * hPtr[10];   // Defines an array of 10 character Pointers
for(int ctr = 0; ctr < 10; ctr++)
{ // Allocates 15 bytes for each hPtr element.
hPtr[ctr] = (char *)malloc(15 sizeof(char)); }

If a user enters more than 15 ( or is it 14) bytes worth what could happen?

Code:

char * hPtr[10];   // Defines an array of 10 character Pointers
for(int ctr = 0; ctr < 10; ctr++)
{ // Allocates 15 bytes for each hPtr element.
hPtr[ctr] = (char *)malloc(15 sizeof(char)); }

should be

Code:

char * hPtr[10];
for(int ctr = 0; ctr < 10; ctr++){
    /*Allocates space for 15 characters to each pointer*/
    hPtr[ctr] = malloc(15 * sizeof(char));
}

-Prelude

As for your second question, if the user enters more characters than allowed for the 15 will be entered and the rest thrown away. If the user enters less than allocated, the array will still have some room left but nothing dire will happen when you access it.

-Prelude

> stating that it couldn't convert a void pointer to a charater pointer
That's because you're compiling C code with a C++ compiler.

Make sure your source code is .c, not .cpp or whatever your compiler treats c++ source files as.

hPtr[ctr] = (char *)malloc(15 sizeof(char));
hPtr[ctr] = malloc(15 sizeof(char));

Both are equivalent in the fact that they're wrong. The error is in the actual call to malloc
malloc(15 sizeof(char));
15 sizeof(char) will not allocate space for 15 characters, you have to multiply the sizeof char by 15 to do that, hence
malloc(15 * sizeof(char));

The other difference is the first one uses a cast and the second doesn't. With the current standard, a cast is not needed anymore though in the past it was required. So on that the choice is up to you, they're both the same.

-Prelude