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c main
print ouput without using puts or printf in c
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main function in c
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int main() {
print int in c
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print an int C
#include<stdio.h> int main() { int n; scanf("%d",&n); int ar[n]; int count=0; for(int i=0;i<n;i++) { scanf("%d",&ar[i]); } for(int i=0;i<n;i++) { for(int j=0;j<n;j++) { if(ar[i]==ar[j]) { count=count+1; } else { count=0; } } } printf("%d\n",(count)/2); }
#include<stdio.h> main () { int a; printf("Enter the number:"); scanf("%d",&a); printf("the number was:",a); return 0; }
%i in printf in C
#include <stdio.h> int main (void) { int a, b; printf("Dwse mou dio akeraious arithmous: \n"); scanf("%d %d", &a, &b); printf("Oi arithmoi poy edwses einai: %d kai %d", a, b); return 0; }
#include <stdio.h> int main() { char array [100]; scanf("%s", array); printf("%s",array); return 0; }
#include <stdio.h> int main() { // printf() displays the string inside quotation printf("Hello, World!"); return 0; }
#include<stdio.h> int main { int arr[ 5]), i=0; while(i<5) arr[i]=++i; for i=0; i<5; i++) printf("%d", arr[i]; return 0;
what do I return in int main() function c++
#include <stdio.h> int main() { int x = 10, *y, **z; y = &x; z = &y; printf(""%d %d %d"", *y, **z, *(*z)); return 0; }
Predict the output of the given code: void method2(char c) { System.out.println( (int) c); } What will be the output of the given code snippet if i) if 'x' is passed to c i) if ‘0’ is passed to c
#include<stdio.h> main() { int *p = 15; printf("%d",*p); }
#include<stdio.h> int main() { for(5;2;2) printf("Hello"); return 0; }
void main() { int i = 0; if (i == 0) { printf("Hello"); continue; } }
#include <stdio.h> int main() { int x = 10, *y, **z; y = &x; z = &y; printf(""%d %d %d"", *y, **z, *(*z)); return 0; }
#include <stdio.h> int main() { int x = 10, *y, **z; y = &x; z = &y; printf(""%d %d %d"", *y, **z, *(*z)); return 0; }
#include <stdio.h> int main() { int x = 5; if (x = 0) printf("Case 1"); else if(x == 0) printf("Case 2"); else printf("Case 3"); }
#include <stdio.h> int main() { int num,i=1; printf("Enter the number whose table you want to print?"); scanf("%d",&num); table: printf("%d x %d = %d\n",num,i,num*i); i++; if(i<=10) goto table; }
#include <stdio.h> int main() { int num,i=1; printf("Enter the number whose table you want to print?"); scanf("%d",&num); table: printf("%d x %d = %d\n",num,i,num*i); i++; if(i<=10) goto table; }
#include <stdio.h> int main() { int x = 10, *y, **z; y = &x; z = &y; printf(""%d %d %d"", *y, **z, *(*z)); return 0; }
#include <stdio.h> int pgcd(int nbr1, int nbr2) { if (nbr2 != 0) return pgcd(nbr2, nbr1%nbr2); else return nbr1; } int main() { int nbr1, nbr2; printf("Entrez deux entiers: "); scanf("%d %d", &nbr1, &nbr2); printf("PGCD de %d et %d = %d", nbr1, nbr2, pgcd
printing part of a string in c with printf
#include<stdio.h>
int fun( )
{
static int num = 10;
//static int num;
//num = 10;
//static => initialized only once -- when program is started ( like global variables )
num++;
printf("num = %d\n",num);
}
int main( )
{
#include<stdio.h>
static int n;
static int k;
int b[100];
int *corder(int *s)
{
int temp;
// static int k;
b[k]=*s;
k++;
for(int i=0;i<k;i++)
{
if(b[i]>temp)
{
temp=b[i];
printf("%d",temp);
#include<stdio.h>
void main()
{
int arr[50],f=-1,r=-1,n,a,b;
do{
printf("Press 1 for Enqueue\n2 for Peek\n3 for Dequeue\n4 for exit\n");
scanf("%d",&a);
switch (a)
{
case 1:
for(int i=0;i<n;i++){
if(i==0){
prin
#include <stdio.h>
int main() {
int a,b,c;
scanf("%d %d%d", &a, &b ,&c);
if (a == b && b==c) {
printf("No\n");
} else if{a!=b && b!==c && c!==a)
printf("No\n");
}else{
printf("Yes");
}
return 0;
}
fix
%c in printf in c
#include <stdio.h> int main() { int x=(20||40)&&(10); printf("%d",x); return 0; }
%i in printf in C
public class Frazione { public static void main(String[] args) { Frazione a = new Frazione(3, 4); Frazione b = new Frazione(5, 4); System.out.println(getAddizione(a + b)); } }
#include int main(){ int i,j; i=j=2,3; while(--i&&j++) printf(%d %d,i,j); return 0; }
I need to write an int function in which there are only cout statements and if I return 0/1 it prints them too.
int main(int c, char *v[]) { char *c; c=v[1] + v[2] + v[3]; printf(“%c”, c); }