C Programming - Memory Allocation - Discussion
Discussion Forum : Memory Allocation - Find Output of Program (Q.No. 3)
3.
What will be the output of the program?
#include<stdio.h>
#include<string.h>
int main()
{
char *s;
char *fun();
s = fun();
printf("%s\n", s);
return 0;
}
char *fun()
{
char buffer[30];
strcpy(buffer, "RAM");
return (buffer);
}
Answer: Option
Explanation:
The output is unpredictable since buffer is an auto array and will die when the control go back to main. Thus s will be pointing to an array , which not exists.
Discussion:
16 comments Page 2 of 2.
Vishal said:
1 decade ago
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.
To over come the problem, declare character array as 'static char buffer[30]' instead 'char buffer[30]' or declare character array Globally.
Rahul said:
1 decade ago
Take buffer variable as a globle to print RAM.
Shamu said:
1 decade ago
In order to print "RAM" how the program must be modified ?
Ranjit said:
1 decade ago
@Shilpi
Here buffer is basically a pointer, so it can be returned.
Here buffer is basically a pointer, so it can be returned.
Shilpi said:
1 decade ago
But an array cannot exist in a return statement. It will cause an error.
Deepak said:
1 decade ago
May also cause stack smashing
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