Warning: implode(): Invalid arguments passed in /www/wwwroot/jobquiz.info/mdiscuss.php on line 336 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<=4; i++) printf("%s,", names[i]); ?->(Show Answer!)
1. 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<=4; i++) printf("%s,", names[i]); return 0; }
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By: guest on 01 Jun 2017 06.02 pm
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".