Questions for: Const
#include<stdio.h>
const char *fun();
int main()
{
*fun() = 'A';
return 0;
}
const char *fun()
{
return "Hello";
}
#include<stdio.h>
#define MAX 128
int main()
{
char mybuf[] = "India";
char yourbuf[] = "BIX";
char const *ptr = mybuf;
*ptr = 'a';
ptr = yourbuf;
return 0;
}
Step 1: char mybuf[] = "India"; The variable mybuff is declared as an array of characters and initialized with string "India".
Step 2: char yourbuf[] = "BIX"; The variable yourbuf is declared as an array of characters and initialized with string "BIX".
Step 3: char const *ptr = mybuf; Here, ptr is a constant pointer, which points at a char.
The value at which ptr it points is a constant; it will be an error to modify the pointed character; There will not be any error to modify the pointer itself.
Step 4: *ptr = 'a'; Here, we are changing the value of ptr, this will result in the error "cannot modify a const object".
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#include<stdio.h>
int main()
{
const int k=7;
int *const q=&k;
printf("%d", *q);
return 0;
}
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#include<stdio.h>
int main()
{
const int x;
x=128;
printf("%d\n", x);
return 0;
}
A const variable has to be initialized when it is declared. later assigning the value to the const variable will result in an error "Cannot modify the const object".
Hence Option B is correct
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#include<stdio.h>
const char *fun();
int main()
{
char *ptr = fun();
return 0;
}
const char *fun()
{
return "Hello";
}
Discuss About this Question.
Discuss About this Question.