About int and unsigned int-CodePudding

There are the following programs：

#include <stdio.h>

int main(void)
{
   int i=2147483647;
   unsigned int j=4294967295;
   printf("%d %d %d\n",i,i 1,i 2);
   printf("%u %u %u\n",j,j 1,j 2);
   return 0;
}

Why i 2 is not equal to -2147483646 ?

why j 2 is not equal to 2

It's the result that I expected was different. What is its execution process like?

EDIT

The result I get is:

i=2147483647
i 1=-2147483648
i 2=-2147483647
j=4294967295
j 1=0
j 2=1

CodePudding user response：

If you will output the value of j in the hexadecimal notation like for example

unsigned int j = UINT_MAX;
printf( "j = %u, j = %#x\n", j, j );

You will get the following output

j = 4294967295, j = 0xffffffff

So adding 1 to 0xffffffff you will get 0x00000000. Again adding 1 you will get 0x00000001.

From the C Standard (6.2.5 Types)

... A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.

As for the signed integer variable i then in general the result is undefined due to the overflow.

CodePudding user response：

according to your question i think that you know why i 1=-2147483648 so i 2 = 2147483647 2 = -2147483647 it's not -2147483648 2 the same thing about second question