csapp 2.2.5 Signed versus Unsigned in C

2.2.5 Signed versus Unsigned in C

C supports both signed and unsigned arithmetic for all of its integer data types.
most numbers are signed by default.
Adding charchter 'U' or 'u' as a suffix creates an unsigned constant
C allows conversion between unsigned and signed
- most systems follow the rule that the underlying bit representation does not change(这里意思是底层的表示不变，那么怎么表示就是看是有符号还是没有符号)
- unsigned ( ightarrow) signed (U2T_w)
- signed ( ightarrow) unsigned (T2U_w)
  (w is the number of bits for the data type)

Conversions can happen due to explicit(显式) casting:

int tx, ty;
unsigned ux, uy;

tx = (int) ux;
uy = (unsigned) ty;

happen implicitly：

int tx, ty;
unsigned ux, uy;

tx = ux; /* Cast to signed */
uy = ty; /* Cast to unsigned */

print a value of type int with directive %u and a value of type unsigned with directive %d:

int x = –1;
//0x8000 0000
unsigned u = 2147483648; /* 2 to the 31st */

printf(“x = %u = %d
”, x, x);
printf(“u = %u = %d
”, u, u);

//answer
x = 4294967295 = –1
u = 2147483648 = –2147483648

-8的补码1000
signed + unsigned ( ightarrow) unsigned + unsigned (implicity)
example:

#include<bits/stdc++.h>
using namespace std;
int main ()
{
    printf("%d",(-1<0U));
    return 0;
}

Consider the comparison –1 < 0U,the second operand is unsigned.the first one is implicitly cast to unsigned. so the expression is equivalent to the comparison 4294967295U < 0U