C Programming - Structures, Unions, Enums - Discussion

Yeah correct output is.

Sanjay 37.
Compiled on devc++.

Correct output will be-

sanjay,37

explanation::-all declaration is correct ,like-structure declaration & function declaration(with global scope)

when modify() is called "strupr()" converts "sanjay" into "sanjay"

note:-strupr() is function of <string.h> so it must be included before using it..!!

and p->age=p->age+2;
add 2 in 35 (i.e 35+2=37)

This looks wrong on many counts.

Struct says: char name[20], and yet "Sanjay" is of type (const char *). You're assigning a pointer to an array?

Can any one explain clearly?

void modify(struct emp*);
has been declared before struct.
if it z declared after struct then the o/p vl be
sanjay 37

According to explanation...the compiler can't identify the prototype of "struct emp" at the declaration because it's not defined yet. So compiler can't estimate the memory need & tools.

No Error in the program. We are just instructing the compiler about the prototype of function modify. It is the abstraction so no need to estimate the size of struct emp.

You need to put the statement "void modify(struct emp*);" after declaring struct because the prototype tells the compiler that there is a struct which is mentioned before hence leading to an error. So prog will look like this:


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

struct emp
{
char name[20];
int age;
};
void modify(struct emp*);
int main()
{
struct emp e = {"Sanjay", 35};
modify(&e);
printf("%s %d", e.name, e.age);
return 0;
}
void modify(struct emp *p)
{
p ->age=p->age+2;
}

What is meant by prototype?

I agree with rahul there is no error while compiling or executing the program.