from main-memory to standard output.
int main (void) { if(printf("Print whatever you want")) { } }
it's not a statement, it's a function: len= printf (format, ...more-parameters...);
printf does return the length: size_t len = printf ("%s", str);
It is a function.
try to usecondition ? value if true : value if falseor: if (printf ("Hello")) {}
I don't really get your point, it is unlikely that both then-statement and else-statement should be executed. I daresay it is impossible.Try me:printf ("before if\n");if (1) printf ("condition is true (non zero)\n");else printf ("condition is false (zero)\n");printf ("after if\n");
main() { if( !fork() ) printf("Hello"); else printf("World"); } This works fine. if (!printf("hello")) printf("Hello"); else printf(" World\n");
if (a > b && a > c) printf("%d\n", a); else if (b > c) printf("%d\n", b); else printf("%d\n", c);
Answerwell if it is about the include statement only then your answer goes simple. extern "C"{int printf(const char *format,..);}int main(){printf("Hello world");}coz all the include statement does is copy the requested file at the asked location.
No, the printf() statement in C can generate multiple lines of output. You can include newline characters (\n) within the string to create line breaks, allowing for formatted output across multiple lines. Additionally, you can call printf() multiple times to print different lines.
Everywhere in the code when there is no 'break', the running will continue with the next instruction. Example: for (i=0; i<3; ++i) { printf ("i=%d: ", i); switch (i) { case 2: printf ("two "); case 1: printf ("one "); case 0: printf ("zero "); } printf ("\n"); } The output: 0: zero 1: one zero 2: two one zero
int f(int num) { int i; for (i = 0; i < 10; i++) { if (num > i) continue; printf("aaa"); } } printf will only be reached when i <= num.