C将char数组打印为float

Question

I'm trying to print a char array of 4 elements as a float number. The compiler(gcc) won't allow me to write zs={'3','4','j','k'}; in the main() function, why?

#include <stdio.h>

union n{
    char s[4];
    float x;
};
typedef union n N;

int main(void)
{
    N z;
    z.s[0]='3';
    z.s[1]='4';
    z.s[2]='j';
    z.s[3]='k';
    printf("f=%f\n",z.x);
    return 0;
}

The output of the program above is: f=283135145630880207619489792.000000 , a number that is much larger than a float variable can store; the output should be, in scientific notation, 4.1977085E-8 . So what's wrong?

Answer 1

zs={'3','4','j','k'}; would assign one array to another. C doesn't permit that, though you could declare the second and memcpy to the first.

The largest finite value that a single-precision IEEE float can store is 3.4028234 × 10^38, so 283135145630880207619489792.000000, which is approximately 2.8313514 × 10^26 is most definitely in range.

Assuming your chars are otherwise correct, the knee-jerk guess would be that you've got your endianness wrong.

EDIT: 34jk if taken from left to right, as on a big-endian machine is:

0x33 0x34 0x6a 0x6b
= 0011 0011, 0011 0100, 0110 1010, 0110 1011

So:

sign = 0
exponent = 011 0011 0 = 102 (dec), or -25 allowing for offset encoding
mantissa = [1] 011 0100 0110 1010 0110 1011 = 11823723 / (2^23)

So the value would be about 4.2 × 10^-8, which is what you want.

In little endian:

0x6b 0x6a 0x34 0x33 
= 0110 1011, 0110 1010, 0011 0100, 0011 0011

sign = 0
exponent = 110 1011 0 = 214 (dec) => 87
mantissa = [1]110 1010 0011 0100 0011 0011 = 15348787 / (2^23)

So the value would be about 2.8 * 10^26, which is what your program is outputting. It's a safe conclusion you're on a little endian machine.

Summary then: byte order is different between machines. You want to use your bytes the other way around — try kj43 .

Answer 2

您实际看到的是{'k''j''4''3'}