c programming

Question

What is the output of the following C program

void main( )
{
static char a[5][20] = {“Mona”, “Vijay”, “Kumar”, “Ankur”, “Suresh”};
int i;
char *p;
p = a[3];
a[3] = a[4];
a[4] = p;
for(i = 0; i <= 4; i++)
printf(“%s”, a[i]);
}

Answer 1

Array is a constant pointer. So, as a restriction of constant pointer it cannot change it's value like a general pointer.

i.e., we cannot change the value of a constant pointer like

a[3] = a[4];

This will give compile time error.

If we want to write this code without any compile time error it can be done as follows:

#include<stdio.h>

int main(void) {
        static char* a[20] = {"Mona", "Vijay", "Kumar", "Ankur", "Suresh"};
        int i;
        char *temp = a[4];
        a[4] = a[3];
        a[3]=temp;
        for(i = 0; i <= 4; i++)
                printf("%s", a[i]);
        return 0;
}

Comments

@Akriti

char *a[20] is array of pointers , where a has 20 pointers.like this

No it is not a constant pointer now. , as it is not contiguously allocating memory now.

yes a[3] = a[4]  is valid.

because 1D array also a pointer a[3]=*(a+3)

After exchanging pointer it's value will be like that

This routine execution eleminates constant pointer problem.

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@Arjun Sir

why strcpy? value also exchanging. Actually it is out of order execution. So, value exchange is not a problem here. Moreover it eleminates complicated storage problem (like twin problem). So, overhead size reduced. rt?

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@srestha yes, it is correct. I was thinking something else.