[Solved] What will be the output of the program ? #include<stdio.h> int *check(static int, static int); int main() { int *c; c = check(10, 20); printf( %d\n , c); return 0; } int *check(static int i, static int j) { int *p, *q; p = &i; q = &j; if(i >= 45) return (p); else return (q); } - Programming - C Programming - Pointers

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What will be the output of the program ?

#include<stdio.h>
int *check(static int, static int);

int main()
{
    int *c;
    c = check(10, 20);
    printf("%d\n", c);
    return 0;
}
int *check(static int i, static int j)
{
    int *p, *q;
    p = &i;
    q = &j;
    if(i >= 45)
        return (p);
    else
        return (q);
}

A 10

B 20

C Error: Non portable pointer conversion

D Error: cannot use static for function parameters

ANS:A - 10

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What will be the output of the program ?
`#include<stdio.h> int check(static int, static int); int main() { int c; c = check(10, 20); printf("%d\n", c); return 0; } int check(static int i, static int j) { int p, *q; p = &i; q = &j; if(i >= 45) return (p); else return (q); }`

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What will be the output of the program ? #include<stdio.h> int *check(static int, static int); int main() { int *c; c = check(10, 20); printf("%d\n", c); return 0; } int *check(static int i, static int j) { int *p, *q; p = &i; q = &j; if(i >= 45) return (p); else return (q); }

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What will be the output of the program ?
`#include<stdio.h> int check(static int, static int); int main() { int c; c = check(10, 20); printf("%d\n", c); return 0; } int check(static int i, static int j) { int p, *q; p = &i; q = &j; if(i >= 45) return (p); else return (q); }`

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