二进制函数的字符无法正常工作

Question

I have made a function to translate a number to its binary form:

size_t atobin (char n)
{
    size_t bin= 0, pow;
    for (size_t c= 1; n>0; c++)
    {
        pow= 1;
        for (size_t i= 1; i<c; i++) //This loop is for getting the power of 10
            pow*= 10;
        bin+= (n%2)*pow;
        n/= 2;
    }
    return bin;
}

It works great for numbers 1 to 127, but for greater numbers (128 to 255) the result is 0... I've tried using the type long long unsigned int for each variable but the result was the same. Someone has an idea about why?

Answer 1

char by default in C is considered to be of signed.

char is of 8 bits(mostly). And for signed char the MSB is used for sign. As a result you can only use 7 bits.

( 0111 1111 ) ₂ = ( 127 ) ₁₀ The maximum value that your fucntion can work with. (as you are passing a type of variable which can hold 127 at max).

If you use unsigned char then the MSB is not used as sign-bit. All 8 bits are used giving us a maximum possible value ( 1111 1111 ) ₂ = ( 255 ) ₁₀

For signed number min/max value is -127 to +127 .

For unsigned number min/max value is 0 to +255 .

So even if you make the type of the passed parameter unsigned char the maximum value it can hold is +255 .

---

A bit more detail:

Q) What happens when you assign >127 values to your char parameter?

It is signed char by default. It is of 8 bits. But it can't hold it. So what will happen?

The result is implementation defined. But

Suppose the value is 130 . In binary it is 10000010 . In most of the cases this returns -126 . So that will be the value of n . n>0; fails. Loop is never entered. And it returns 0 .

---

Now if we make it unsigned char then it can hold values between 0 and 255 (inclusive). And that is what you want to have here.

---

Note:

Q) What happens when >255 values are stored in unsigned char?

The value is reduced to modulo of (max value unsigned char can hold+1) which is 256.

So apply modulo operation and put the result. That will be stored in unsigned char.