C指针-按值传递

Question

I want to know why is it that when i call readF() and have it return the buffer pointer and i print the buffer in main() it prints the actual data and not the memory address? I am returning the address and not the actual data correct? In my main i have a pointer that points to what gets returned, which is just what the buffer pointer in readF() points to.

char *readF(){

char *buffer=NULL;

//allocate memory to contain the string plus null terminator
buffer=malloc((sizeof(char)*4)+1);
//fill memory with string Hello plus terminator

return buffer;
}

int main(int argc, char **argv){
char *buffer;
buffer=readF();
printf("%s", buffer);

return 0;
}

Answer 1

You print the actual data inside the buffer because you use the %s format specified in your printf statement. This tells printf to assume the pointer it receives, buffer in this case for the first specifier, is a pointer to a NULL-terminated character string, so it prints the string.

If you want to print the address use the %p format specifier

You're correct when you say the readF function is returning the memory address. It returns the memory address of the first byte of the buffer, ie a pointer to the buffer you allocate.

Note: at the end of your function you should really free(buffer) too :)

Answer 2

To print the address you need %p specifier.
Change

printf("%s", buffer);  

to

printf("%p", (void *)buffer);