A written question by Dahua Corporation C/C ++

After the program is debugged, the result is 5,-7,-2;

#include"stdio.h"#include"iostream"using namespacestd;int main(void){      unsigned int a=5;        int b=-7,c=4;        if(b+a>0)               c=a+b,b=c-a;        else               c=a-b,b=c+a;        cout<<a<<","<<b<<", "<<c<<endl;                return 0;}

The result is c = 12, B = 17, and a = 5. This is mainly because B + A equals-2 is less than zero and is executed in else. After one-step debugging in Visual C ++, it is found that the program enters the if statement after the if statement is determined. That is, B + A is greater than zero. Why didn't I enter else? After looking at the code again, we know that the unsignedint type is used for variable A, while B is of the int type. What type is the result after the sum of the two? This involves a problem of upgrading the type when calculating different types of variables!

When two variables of different types are added, the result type should follow the principle below: In the added variables, the result is converted to which type of variable can represent a larger value. SHORT + long, the result is converted to long. The sum of unsigned int and INT types. on 32-bit machines, the maximum value of unsigned int can be 2 ^ 32-1, and the maximum value of int can be 2 ^ 31-1. because the result should be converted to the unsignedint type. In this way, int is upgraded to unsignedint.

Unsigned int A = 5; int B =-7; B is a negative value, and the highest bit is 1. After being converted to an unsigned integer, the highest bit does not represent a symbol, but a value of this type, is a very large positive number, and the result of adding both is also a large integer. So it is natural to enter the if statement.

In the IF and else branches, a comma expression is used, for example:

C = A + B, B = C-A; execute c = a + B first, and then execute B = C-a. the final value of the entire expression is the value of the last expression, if int resullt = (C = A + B, B = C-A), the result is the same as the value of B.

Binary representation:

Unsignedint A = 5; red

Int B =-7; black

Makeup: 00000000 00000000 00000000 00000101

Original code: 10000000 00000000 0000000 00000111

Makeup: 111111111 11111111 11111111 11111001

Complement: 11111111 11111111 11111111 11111110

Converts data to C, where C is an int;

The highest bit is 1, which is a negative number. The complement code is converted to the original code.

First subtract multiple reverse attempts

11111111 11111111 11111111 111111101

10000000 00000000 00000000 00000010

-2 is the result of C.