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answer #include<stdio.h> main() { int *p = 15; printf("%d",*p); }

related #include <stdio.h> int main() { int x = 10, *y, **z; y = &x; z = &y; printf(""%d %d %d"", *y, **z, *(*z)); return 0; }

related #include<stdio.h> int main() { for(5;2;2) printf("Hello"); return 0; }

related #include <stdio.h> int main() { int x = 10, *y, **z; y = &x; z = &y; printf(""%d %d %d"", *y, **z, *(*z)); return 0; }

related #include <stdio.h> int main() { int x = 10, *y, **z; y = &x; z = &y; printf(""%d %d %d"", *y, **z, *(*z)); return 0; }

related #include <stdio.h> int main() { int x = 10, *y, **z; y = &x; z = &y; printf(""%d %d %d"", *y, **z, *(*z)); return 0; }

related #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

related #include <stdio.h> int main() { int x=(20||40)&&(10); printf("%d",x); return 0; }

More Related Answers

#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; }

What will be output when you will execute following c code? #include<stdio.h> int main(){ printf("%d\t",sizeof(6.5)); printf("%d\t",sizeof(90000)); printf("%d",sizeof('A')); return 0;

#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 will be output when you will execute following c code? #include<stdio.h> int main(){ printf("%d\t",sizeof(6.5)); printf("%d\t",sizeof(90000)); printf("%d",sizeof('A')); 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 <iostream> #include <stdio.h> using namespace std; int main() { int a[5]; a[0]=12; a[1]=13; a[2]=14; a[3]=15;

#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() { //output a line printf("Hello world!"); }

#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

#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);

int main(int c, char *v[]) { char *c; c=v[1] + v[2] + v[3]; printf(“%c”, c); }