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

You are using a wrong format specifier for `uint32_t`. See https://stackoverflow.com/questions/3168275/printf-format-specifiers-for-uint32-t-and-size-t — Eugene Sh., Nov 25 '21 at 21:29
`printf()` doesn't know or care what data was passed to it (or even if there was none passed). If you tell it `%d` then it will look where it expects the data to be and interpret it as if it is `int` type, no matter what is passed. If a floating point variable was passed, unlike normal functions there is no type conversion (apart from type promotions) and again `printf()` looks at 32-bit data and assumes it is `int` data. — Weather Vane, Nov 25 '21 at 21:39

score 4 · Answer 1 · answered Nov 25 '21 at 21:31

4

%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.

answered Nov 25 '21 at 21:31

dbush

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There could be a pathological case where `unsigned int` is smaller than `uint32_t`. Then this won't work. There is a "proper" format specifier `"%"PRIu32` for this – Eugene Sh. Nov 25 '21 at 21:32
thanks this explains the printf problem. But it doesnt explain why it says ffffff4 is lower than fffffff2 – Magnus Mikkelsen Nov 25 '21 at 21:36
Because `-14` is smaller then `-12` – KamilCuk Nov 25 '21 at 21:38
@MagnusMikkelsen `0xfffffff4` is unsigned and does not fit into a signed `int`. So it is being converted with implementation-defined result (as per http://port70.net/~nsz/c/c11/n1570.html#6.3.1.3p3). In your case it is most likely interpreting the bit patter as 2's complement represented negative number. – Eugene Sh. Nov 25 '21 at 21:38
test2 uses unsigned integers so that should give me back 1 as 0xfffffff4 is higher than fffffff2 – Magnus Mikkelsen Nov 25 '21 at 22:58

score 1 · Answer 2 · edited Nov 25 '21 at 23:21

Use the correct formats for fixed size integers:

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

 void main() {
    int32_t reg1S = 0xfffffff4; \\ <- implementation defined
    int32_t reg2S = 0xfffffff2; \\ <- implementation defined
          
    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);
 }

uint32_t is having a negative value when MSB is 1

2 Answers2