Question

The code below ends up in a seemingly endless loop while printing some decimal numbers.

int main(){
    show(0xBADECAFU);
}

void show(unsigned a){
    unsigned pos=0;
    for(; pos<UINT_MAX; pos++)
        printf("%u", (1U<<pos) & a);
}

The code below actually shows the bits of the hex number. Why does the first program run improperly while the second does not?

int main(){
     show(0xBADECAFU);
}

void show(unsigned n){
    unsigned pos=31, count=1;
    for(; pos!=UINT_MAX; pos--, count++){
        printf("%u", n>>pos & 1U);
}

Answer 1

There are not UINT_MAX bits in an unsigned int. There are, however, CHAR_BIT * sizeof(unsigned int) bits.

/* nb: this prints bits LSB first */
void show(unsigned a){
    unsigned pos=0;
    for(; pos < CHAR_BIT*sizeof(unsigned); pos++)
        printf("%u", (1U<<pos) & a ? 1 : 0);
}

Consider your second case, where you loop until pos equals UINT_MAX. This will properly^* print out 32 bits of unsigned, assuming underflow goes to ~0 and sizeof(unsigned) is at least 4.

Your second example could be improved slightly:

void show(unsigned n){
    int pos = (CHAR_BIT * sizeof(unsigned)) - 1;
    for(; pos >= 0; pos--) {
        printf("%u", (n>>pos) & 1U);
    }
}

_{* Your code which “prints” the bits was odd, and in my example I’ve fixed it up.}