在printf上输出

Question

is there anyone who can give me a vital and simple explanation im trying to understand this program below its about the printf in c programming here is the code......

#include <stdio.h>
  int main()
   {
    int a,b;
    float c;
      a = 1;
      b = 2;
      c = 3.;

    printf("%d %d %f\n",a ,b , c);
    printf("%d %f %f\n",a ,b , c);
    printf("%f %d %f\n",a ,b , c);
    printf("%d %d %d\n",a ,b , c);
    printf("%f %d %d\n",a ,b , c);
   return o;

    }

the output is

1  2  3.000000
1  0.000000  0.000000
0.000000  0  0.000000
1  2  0
0.000000  0  1074266112

im perfectly clear about the first printf's output but why does the second printf outputs 1 0.000000 0.000000

instead of 1 0.000000 3.000000

third printf 0.000000 0 3.000000

fifth printf 0.000000 2 0.000000

Answer 1

The behaviour of all the printf() statements except the first one is undefined . You therefore cannot expect them to behave in any particular way.

To fix this, you need to cast the numeric arguments to printf() so that their types match the format specifier. The compiler does not perform this conversion automatically, although some good compilers would issue a warning to alert you to the mistake.

Answer 2

When you pass an int to printf() and tell it to treat it as a float , you get the garbage out corresponding to the garbage in (GIGO). Basically, neither printf() nor the compiler can do any conversions for you. You have to get it right. You must supply arguments that match the format specification. These comments apply to all the printf() and scanf() families of functions.

If you use GCC and string literals as the format strings (as in your example), you will get compiler warnings about type mismatches. Pay heed and fix them; the compiler knows more than you do. If you're using a different compiler that does not give such warnings, find one (such as GCC) that does - and use it. And make sure you turn the warnings on - and that your code compiles without emitting any warnings.

You are invoking 'undefined behaviour' with the second and subsequent statements, which means any and all answers (and other behaviours, including crashes and reformatting your disk) are acceptable responses by the system.

Answer 3

You're running into undefined behavior . The runtime expects a float and you give it an int .

Answer 4

To expand above answers, and give a clue as to what is causing the undefined behavior. On your platform, floats have a different size than ints, so printf() is wrong not only about the type of its arguments, but also about where they are in memory. So in the second line, printf is not looking at the start of the float when it prints the third value; it is looking somewhere else.