How does C store negative numbers in signed vs unsigned integers?

Question

Here is the example:

#include <stdio.h>

int main()
{
    int x=35;
    int y=-35;
    unsigned int z=35;
    unsigned int p=-35;
    signed int q=-35;
    printf("Int(35d)=%d\n\
Int(-35d)=%d\n\
UInt(35u)=%u\n\
UInt(-35u)=%u\n\
UInt(-35d)=%d\n\
SInt(-35u)=%u\n",x,y,z,p,p,q);

    return 0;
}

Output:

Int(35d)=35
Int(-35d)=-35
UInt(35u)=35
UInt(-35u)=4294967261
UInt(-35d)=-35
SInt(-35u)=4294967261

Does it really matter if I declare the value as signed or unsigned int? Because, C actually only cares about how I read the value from memory. Please help me understand this and I hope you prove me wrong.

Answer 1

Does it really matter if I declare the value as signed or unsigned int?

Yes.

For example, have a look at

#include <stdio.h>

int main()
{
    int a = -4;
    int b = -3;
    unsigned int c = -4;
    unsigned int d = -3;
    printf("%f\n%f\n%f\n%f\n", 1.0 * a/b, 1.0 * c/d, 1.0*a/d, 1.*c/b);
}

and its output

1.333333
1.000000
-0.000000
-1431655764.000000

which clearly shows that it makes a huge difference if I have the same byte representation interpreted as signed or unsigned.