I have a query about using the memcpy() function.I have written the below program, it compiles but doesn't print an output. The .exe opens and the crashes. I am using Code Blocks as my IDE using GNU GCC compiler.
int main()
{
char a[]= "This is my test";
char *b;
memcpy(b,a,strlen(a)+1);
printf("After copy =%s\n",b);
return(0);
}
However, if I change the array *b to b[50] it works!! I don't understand why.
Please provide your suggestions!
Thanks!
Your pointer b is uninitialized. It is pointing to some random location in memory. So when you copy stuff into the memory which b is pointing to, bad things are likely to happen.
You need to initialize it; perhaps allocate some memory for it with malloc().
char *b = malloc(strlen(a) + 1);
And then free it when you're finished.
free(b);
You are lucky it did not crash when you used pointer - it should have.
When you copy memory, destination must be allocated first. If you use char b[50], you allocate 50 bytes for b on stack. If you use char *b, you did not allocate anything yet, and typically should do this using something like malloc : b = malloc(50);.
With malloc it will work, but then you should not forget to release that memory with free(b);.
If memory was allocated on stack, release happens automatically.
b[50]man malloc, man free, man memcpy
bpoints to - which is unknown.bis unitialised - it could be anywhere, and there is absolutely no reason to think your program is allowed to write to that memory location.bis not a single char, it's a pointer.