use cout << simple
int main (void) { if(printf("Print whatever you want")) { } }
If you forget the semicolon, your program won't compile.
try to usecondition ? value if true : value if falseor: if (printf ("Hello")) {}
printf("%u %+d %+d",200,200,-200); //will print 200 +200 -200
With functions like ecvt, fcvt, gcvt.
you can put the printf function inside a if() so that you need not use semicolons. for eg to print my name, if(printf("My Name is Maheedharan")) {} this will work. try this out...
#include <stdio.h> int main (void) { puts ("the name"); return 0; }
printf(format_string, ...);
Printf function is used in c language. Printf is used to print something to the standard output. ex: printf ('welcome');
Your question makes no sense.
write, putchar, putc, fputc etc
#include<stdio.h> #include<conio.h> void main() { int i, a[100]; clrscr(); printf("Numbers from 1 to 100"); for(i=1;i<=100;i++) printf(" %d",a[i]); printf("Numbers from 1 to 100 without multiples of 2"); for(i=1;i<=100;i++) { if (a[i]%2 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 3"); for(i=1;i<=100;i++) { if (a[i]%3 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 4"); for(i=1;i<=100;i++) { if (a[i]%4 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 5"); for(i=1;i<=100;i++) { if (a[i]%5 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 6"); for(i=1;i<=100;i++) { if (a[i]%6 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 7"); for(i=1;i<=100;i++) { if (a[i]%7 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 8"); for(i=1;i<=100;i++) { if (a[i]%8 !=0) printf(" %d",a[i]); } printf("Numbers from 1 to 100 without multiples of 9"); for(i=1;i<=100;i++) { if (a[i]%9 !=0) printf(" %d",a[i]); } getch(); }