C Programming - Memory Allocation - Discussion

The output is Garbage value. Because in the function definition, the character array 'buffer[30]' is declared as local. The scope of the local variable is limited within the blocks. But in the example, we are trying to return the address of a local array. This leads to problem & it is known as 'Dangling Pointer'.

To over come the problem, declare character array as 'static char buffer[30]' instead 'char buffer[30]' or declare character array Globally.

The return will be "RAM". The reason is very simple.

1. Return value will be copied to another place in stack.
2. Buffer will be free at the end of fun().

This concept is similar to return a local object in Cpp. The object will be copied to another place in a stack and the local object will be freed at the end of the function.

Below code will return ram.

#include<stdio.h>
#include<string.h>

int main()
{
char *s;
char *fun();
s = fun();
printf("%s\n", s);
return 0;
}
char *fun()
{
static char buffer[30];
strcpy(buffer, "RAM");
return (char*)buffer;
}

DEVC++ is giving the answer RAM. I don't know why and also there is no error because of returning the address of local variable that is buffer, it is just giving the warning but answer is coming.

char *fun()
{
char *buffer;
buffer=(char*)malloc(sizeof(char*));
strcpy(buffer, "RAM");
return (buffer);
}

This should work.

I execute this code and it returns RAM. It is strange. I don't understand why.

Buffer array is already a char what is the need to typecast in last statement?

But an array cannot exist in a return statement. It will cause an error.

@Shilpi

Here buffer is basically a pointer, so it can be returned.

In order to print "RAM" how the program must be modified ?