Question
#include<stdio.h>
int main()
{
char *names[] = { "Suresh", "Siva", "Sona", "Baiju", "Ritu"};
int i;
char *t;
t = names[3];
names[3] = names[4];
names[4] = t;
for(i=0; i
What will be the output of the program ?
#include<stdio.h>
int main()
{
char *names[] = { "Suresh", "Siva", "Sona", "Baiju", "Ritu"};
int i;
char *t;
t = names[3];
names[3] = names[4];
names[4] = t;
for(i=0; i
Answer: Option B
Was this answer helpful ?
Step 1: char *names[] = { "Suresh", "Siva", "Sona", "Baiju", "Ritu"}; The variable names
is declared as an pointer to a array of strings.
Step 2: int i; The variable i is declared as an integer type.
Step 3: char *t; The variable t is declared as pointer to a string.
Step 4: t = names[3]; names[3] = names[4]; names[4] = t; These statements the swaps
the 4 and 5 element of the array names.
Step 5: for(i=0; i<=4; i++) printf("%s,", names[i]); These statement prints the all the value
of the array names.
Hence the output of the program is "Suresh, Siva, Sona, Ritu, Baiju".
Was this answer helpful ?
Submit Solution