#include <stdio.h> #include <stdlib.h> int main() { int x = 1; printf("please make a selection with your keyboard\n"); sleep(1); printf("1.\n"); char input; scanf("%c", &input); switch (input) { case '1': x = x + 1; printf(x); } return(0); } I am trying a make a variable add to itself and then print that variable out, but I can't seem to get my code to work.
My output error is
newcode1.c: In function ‘main’: newcode1.c:20:2: warning: passing argument 1 of ‘printf’ makes pointer from integer without a cast [enabled by default] In file included from newcode1.c:1:0: /usr/include/stdio.h:362:12: note: expected ‘const char * __restrict__’ but argument is of type ‘int’ newcode1.c:20:2: warning: format not a string literal and no format arguments [-Wformat-security] 2 Answers
Your printf needs a format string:
printf("%d\n", x); This reference page gives details on how to use printf and related functions.
As Shafik already wrote, you need to use the right format because scanf gets you a char.
Don't hesitate to look at printf - C++ Reference if you aren't sure about the usage.
Hint: It's faster/nicer to write x = x + 1; the shorter way is: x++;
