What is the output of this program?
#include <stdio.h>
int main() {
printf("%d", printf("ABCD"));
return 0;
}
#include <stdio.h>
int main() {
printf("%d", printf("ABCD"));
return 0;
}
Consider the following C program:
#include <stdio.h>
int main( )
{
int i, j, k = 0;
j=2*3/4+2.0/5+8/5;
k -= --j;
for (i=0; i<5; i++)
{
switch(i+k)
{
case 1:
case 2: printf("\n%d", i+k);
case 3: printf("\n%d", i+k)
default: printf("\n%d", i+k);
}
}
return 0;
}
#include <stdio.h>
int main( )
{
int i, j, k = 0;
j=2*3/4+2.0/5+8/5;
k -= --j;
for (i=0; i<5; i++)
{
switch(i+k)
{
case 1:
case 2: printf("\n%d", i+k);
case 3: printf("\n%d", i+k)
default: printf("\n%d", i+k);
}
}
return 0;
}
#include <stdio.h>
int main()
{
int i;
for(i=1; i<=10; i++){
if(i>5){
break;
}
}
printf("%d",i);
return 0;
}
int main()
{
int i;
for(i=1; i<=10; i++){
if(i>5){
break;
}
}
printf("%d",i);
return 0;
}
Consider the C Programming
#include<stdio.h>
void print (int n) {
if (n <= 0) return;
print(n--);
printf ("%d", n);
}
int main() {
print(5);
return 0;
}
#include<stdio.h>
void print (int n) {
if (n <= 0) return;
print(n--);
printf ("%d", n);
}
int main() {
print(5);
return 0;
}