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

Please explain this a bit more!!!

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@srestha semantic analyser reports this error. rt??

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It has correct token .So, it will not cause Lexical error, as it has correct lexemes. It will not cause syntax error, as there is nothing like';' missing. And even no semantic error [This will happen if one space we use one symol, and where we are using other symbol, like declaring  a variable as int  but using it as char ]

So, this error should be at time of intermediate code generation. When we are trying to make an array as normal pointer

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After removing static it is giving error:

 In function 'main':
8:6: error: incompatible types when assigning to type 'char[20]' from type 'char *'
 a[3] = a[4];
      ^
9:6: error: incompatible types when assigning to type 'char[20]' from type 'char *'
 a[4] = p;
      ^

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No , it will not giving error.

try for any online compiler.

@Arjun Sir plz. tell what is actual work of static 'keyword' here.

I cannot identify it

as array is already a static pointer, so what the purpose of using external static keyword here?

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First array is not a static pointer- it is a const pointer- meaning the value assigned to the array (not array element) cannot be changed.

And in this code there is no particular use of using "static" for 'a'. Usually static in C language is used to make a variable alive across function calls (from start till the end of process). It is like a global variable except that scope is restricted to the enclosing block.

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yes,thanks.

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@srestha final code is correct, but what is it doing?

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Nothing, just pointer exchange . but not value exchange.

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yes, so doing no job. The question wants to modify the 4th and 5th elements of the array. We cannot do this without doing a character by character copy of the strings (strcpy).

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@srestha,as u said that array is a constant pointer,so value pointed to it cant be changed..so in your modified code, "char *a[20]".....here what is *a[20] now...??

is it not a constant ptr now..??

or is it array of 'character pointers' or array of characters which return a pointer to character?

what has changed now so that a[3] = a[4] is valid now?

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here,a[3] gives the base address of Ankur and a[4] gives base address of Suresh.

so a[3]=a[4] is invalid because they both are addresses..right?

but somehting like a[3][2] = a[4][1] is allowed because now we are working on values..right?

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@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.