C programming - unsigned int overflow problem

Question

I want to print out an integer type variable which value is 3000000000;

So I wrote the below code then ran it, but the printed value was incorrect. I think the variable is overflowed. But I don't know why.

#include<stdio.h>

int main(void) {
    unsigned int num1 = 3000000000;

    printf("%d", num1);
}

As far as I know, the maximum value of unsigned integer type variable is (2^32-1 = 4,294,967,296 - 1) when the code complies on Win32 API. But the printed value is -1294967296.

I have no idea why overflow occurs in my code.

If anyone knows the reason, please let me know :)

Best regards,

I use Microsoft Visual Studio 2015 Professional.

%d is for signed ints. It will take whatever value it finds and interpret it as signed int and print it. Try unsigned int num1 = 0xffffffff; printf("%d", num1);. It will be more obvious. 4294967295 = 0xffffffff — Jabberwocky
– Jabberwocky, Commented Mar 11, 2019 at 7:26

Yunbin Liu · Accepted Answer · 2019-03-11 09:31:13Z

use %u not %d

For printf:

%d is used by:

d, i The int argument is converted to signed decimal notation.The precision, if any, gives the minimum number of digits that must appear; if the converted value requires fewer digits, it is padded on the left with zeros. The default precision is 1. When 0 is printed with an explicit precision 0, the output is empty.

%u is used by:

o, u, x, X The unsigned int argument is converted to unsigned octal (o), unsigned decimal (u), or unsigned hexadecimal (x and X) nota‐ tion. The letters abcdef are used for x conversions; the let‐ ters ABCDEF are used for X conversions. The precision, if any, gives the minimum number of digits that must appear; if the converted value requires fewer digits, it is padded on the left with zeros. The default precision is 1. When 0 is printed with an explicit precision 0, the output is empty.

see http://man7.org/linux/man-pages/man3/printf.3.html

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