uint32_t is having a negative value when MSB is 1

Question

I need to make a signed value into an unsigned value and while trying to get this to work I found uint32_t is somehow signed when i assign the MSB to 1.

In this code i show what i mean. Test1/2 are to show this isn't only a printf problem, but arimethics aswell.

    #include <stdio.h>
    #include <inttypes.h>

     void main() {
        int reg1S = 0xfffffff4;
        int reg2S = 0xfffffff2;
              
        uint32_t reg1U = 0xfffffff4;
        uint32_t reg2U = 0xfffffff2;

        uint8_t test1,test2;

        test1 = (reg1S < reg2S ? 1:0);                      
        test2 = (reg1U < reg2U ? 1:0); 

        printf("signed = %d \t",test1);
        printf("unsigned = %d \n", test2);
        printf("reg1S= %d \t", reg1S);
        printf("reg1U= %d \n", reg1U);
        printf("reg2S= %d \t", reg2S);
        printf("reg2U= %d \n", reg2U);
     }

this is what the code outputs

signed = 0  unsigned = 0 
reg1S= -12  reg1U= -12 
reg2S= -14  reg2U= -14

note: it doesnt do this with 8 or 16 bit unsigned ints, only 32 and 64 bit

Answer 1

%d is for printing signed types. To print unsigned types, use %u .

Also, types smaller than int will be promoted to that type when passed to printf which is why you don't see this behavior with those types.

Answer 2

Use the correct formats for fixed size integers:

#include <stdio.h>
#include <inttypes.h>

 void main() {
    int32_t reg1S = 0xfffffff4;
    int32_t reg2S = 0xfffffff2;
          
    uint32_t reg1U = 0xfffffff4;
    uint32_t reg2U = 0xfffffff2;

    printf("reg1S= %"PRIi32" \t", reg1S);
    printf("reg1U= %"PRIu32" \n", reg1U);
    printf("reg2S= %"PRIi32" \t", reg2S);
    printf("reg2U= %"PRIu32" \n", reg2U);
 }