Why is printf printing wrong values?

Question

Why am I getting the wrong values ​​when printing intwith printf("%f\n", myNumber)?

I don’t understand why it prints fine with %d, but not with %f. Shouldn't he add extra zeros?

int a = 1;
int b = 10;
int c = 100;
int d = 1000;
int e = 10000;

printf("%d %d %d %d %d\n", a, b, c, d, e);   //prints fine
printf("%f %f %f %f %f\n", a, b, c, d, e);   //prints weird stuff

Answer 1

Well, of course, he prints "weird" things. You pass in ints but report printfwhich you passed in floats. Since these two types of data have different and incompatible internal representations, you will get "nonsense."

, ​​ printf, " ", , ( ).

, , :

union {
    int n;
    float f;
} x;

x.n = 10;

printf("%f\n", x.f); /* pass in the binary representation for 10, 
                        but treat that same bit pattern as a float, 
                        even though they are incompatible */

Answer 2

float, float, printf.

printf("%f %f %f %f %f\n", (float)a, (float)b, (float)c, (float)d, (float)e);

Answer 3

a, b, c, d e . printf() float, .

Answer 4

printf() Undefined Behaviour

:

 int n=1;
 printf("%f", n); //UB

 float x=1.2f;
 printf("%d", x); //UB

 double y=12.34;
 printf("%lf",y); //UB 

: double printf() %f.

Answer 5

... printf.

if ("%f") {
 float *p = (float*) &a;
 output *p;  //err because binary representation is different for float and int
}

Answer 6

printf , , . "% f% f" printf , , . , .

stdarg.h ,

Answer 7

"" ( ) .

This does not happen with variadac arguments, which are always passed as-is.