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#include <stdio.h> int main() { char arr[] = {“JAYPEE”, “JIIT”, “XML”, “JSON”, “SAM”, “SAMISHTI”}; call(arr); return 0; } void call(char ptr) { char ptr1; ptr1 = (ptr+ = sizeof (int)) – 2; printf(“%s\n”, ptr1);

Poised Peafowl answered on October 20, 2021 Popularity 1/10 Helpfulness 3/10

answer #include <stdio.h> int main() { char *arr[] = {“JAYPEE”, “JIIT”, “XML”, “JSON”, “SAM”, “SAMISHTI”}; call(arr); return 0; } void call(char **ptr) { char ** ptr1; ptr1 = (ptr+ = sizeof (int)) – 2; printf(“%s\n”, *ptr1);

#include <stdio.h> int main() { char arr[] = {“JAYPEE”, “JIIT”, “XML”, “JSON”, “SAM”, “SAMISHTI”}; call(arr); return 0; } void call(char ptr) { char ptr1; ptr1 = (ptr+ = sizeof (int)) – 2; printf(“%s\n”, ptr1);

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#include <stdio.h>
int main()
{ 
char *arr[] = {“JAYPEE”, “JIIT”, “XML”, “JSON”, “SAM”, “SAMISHTI”};
call(arr);
return 0;
}
void call(char **ptr)
{
char ** ptr1;
ptr1 = (ptr+ = sizeof (int)) – 2;
printf(“%s\n”, *ptr1);

Popularity 1/10 Helpfulness 3/10 Language whatever

Source: Grepper

Tags: char int return sizeof void whatever

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Contributed on Oct 20 2021

Poised Peafowl

0 Answers Avg Quality 2/10

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#include <stdio.h> int main() { char arr[] = {“JAYPEE”, “JIIT”, “XML”, “JSON”, “SAM”, “SAMISHTI”}; call(arr); return 0; } void call(char ptr) { char ptr1; ptr1 = (ptr+ = sizeof (int)) – 2; printf(“%s\n”, ptr1);

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#include <stdio.h> int main() { char arr[] = {“JAYPEE”, “JIIT”, “XML”, “JSON”, “SAM”, “SAMISHTI”}; call(arr); return 0; } void call(char ptr) { char ptr1; ptr1 = (ptr+ = sizeof (int)) – 2; printf(“%s\n”, ptr1);

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