How is conversion of float/double to int handled in printf?
c
Solution
The `printf` function uses the format specifiers to figure out what to pop off the stack. So when it sees `%d`, it pops off 4 bytes and interprets them as an `int`, which is wrong (the binary representation of `(float)3.0` is not the same as `(int)3`).
You'll need to either use the `%f` format specifiers or cast the arguments to `int`. If you're using a new enough version of `gcc`, then turning on stronger warnings catches this sort of error:
$ gcc -Wall -Werror test.c
cc1: warnings being treated as errors
test.c: In function ‘main’:
test.c:10: error: implicit declaration of function ‘printf’
test.c:10: error: incompatible implicit declaration of built-in function ‘printf’
test.c:10: error: format ‘%d’ expects type ‘int’, but argument 4 has type ‘double’
test.c:10: error: format ‘%d’ expects type ‘int’, but argument 5 has type ‘double’
Response to the edited part of the question:
C's integer promotion rules say that all types smaller than `int` get promoted to `int` when passed as a vararg. So in your case, the `'d'` is getting promoted to an `int`, then printf is popping off an `int` and casting to a `char`. The best reference I could find for this behavior was this blog entry.
Problem
Consider this program ``` int main() { float f = 11.22; double d = 44.55; int i,j; i = f; //cast float to int j = d; //cast double to int printf("i = %d, j = %d, f = %d, d = %d", i,j,f,d); //This prints the following: // i = 11, j = 44, f = -536870912, d = 1076261027 return 0; } ``` Can someone explain why the casting from double/float to int works correctly in the first case, and does not work when done in printf? This program was compiled on gcc-4.1.2 on 32-bit linux machine. EDIT: Zach's answer seems logical, i.e. use of format specifiers to figure out what to pop off the stack. However then consider this follow up question: ``` int main() { char c = 'd'; // sizeof c is 1, however sizeof character literal // 'd' is equal to sizeof(int) in ANSI C printf("lit = %c, lit = %d , c = %c, c = %d", 'd', 'd', c, c); //this prints: lit = d, lit = 100 , c = d, c = 100 //how does printf here pop off the right number of bytes even when //the size represented by format specifiers doesn't actually match //the size of the passed arguments(char(1 byte) & char_literal(4 bytes)) return 0; } ``` How does this work?