[C programming tutorial experiment] Basic practice questions (2) The error-prone questions need to be reviewed multiple times

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Code 1: - - operator || operator

The output of running the following program is:

//运行以下程序，输出结果是？ 
#include <stdio.h>
int main()
{
    
    
	int a,b,c;
	a=1; b=2; c=3; 
	a=b--<=a||a+b!=c;//首先a取值应当为1或者0。
	//b--<=a 取值应为假，故为0；（2<1显然为假，）
	//a+b!=c 取值应为真，故为1；（1+1！=3显然对√）0||1 =1，故a=1。 
	printf("a=%d,b=%d\n",a,b); //故输出 a=1,b=1。 
}

output:

The reason is explained in the comments!
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Code 2: Mandatory type conversion such as (int) (5.0/2), priority understanding

//运行以下程序的输出结果是? a=3 
#include <stdio.h>
int main()
{
    
    
	int a;
	a=(int)((double)(3/2)+0.5+(int)1.99*2);
	//首先a是整型的，所以0.5应该取整为0，又(double)(3/2)=1.500000,
	
	//对于(int)1.99*2,应该理解为，1.99取整为1，然后1*2=2
	
	//所以 a=1.5+2=3.500000，取整，所以a=3 
	printf("a=%d\n\n",a);  //a=3
	
	printf("----------------\n\n");
	printf("测试_1：(int)1.99*2 = %d\n\n",(int)1.99*2); //2
	printf("测试_1：(int)(1.99*2) = %d\n\n",(int)(1.99*2)); //3
	
	//由测试结果可知,(int)1.99*2 确实应该是先对1.99取整为1，再 1*2=2 
}

output:

The comments are very clear!
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Code 3: Coercion type conversion operator

//求以下表达式的值 a=4
#include <stdio.h>
int main()
{
    
    
	int a;
	a=(int)((double)(5/2) + 2.5);
	//首先， (double)(5/2)=2.000000。因为5/2=2，又double，所以为2.000000
	//其次，2+2.5=4.5。取整，故a=4 
	printf("a=%d\n",a); // a=4
 }

output:
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Code 4: Self-addition operators ++i and i++

//运行以下程序的输出结果是
#include <stdio.h>
int main()
{
    
    
	int i=8,j=9,m,n;
	m=++i; //m=9 i=9
	n=j++; //n=9 j=10
	printf("i=%d,j=%d,m=%d,n=%d\n",i,j,m,n);
	//故 i=9,j=10,m=9,n=9 
	
 }

output:
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Code 5: Priority question (error-prone questions need to be reviewed multiple times)

 //易错题 
//运行以下程序的输出结果为： i=1,j=2,k=0 
#include <stdio.h>
int main()
{
    
    
	int k,i=0,j=2;
	k=i++ && j++ ; 

	//错误思路×××： 0 && 2,结果为假 取0,故k=0。之后,i和j再分别自加，故 i=1,j=3 
	//以上错误的原因是，把最终结果当成了 "k=多少？"
	 
	//正解√：&&作为分隔符，先算左边的，即先是k=i++,此时k=0,i=1,&&左边取值为0
	//由于 && 运算符，一假则假，故 (k=i++ && j++)取值即为0,
	//即 && 右边不执行，所以j仍然为原值，即j=2 
	printf("答案：i=%d,j=%d,k=%d\n\n",i,j,k);
	
	printf("------------------------------\n\n");
	//为便于测试，故重新定义变量q相当于k，同时i和j赋回原来的值。 
	int q; i=0,j=2; 
	
	printf("测试：\n\n");

	printf("(q=i++) && (j++)  = %d\n\n",(q=i++) && (j++)); //0
	
	printf("------------------------------\n\n");
	printf("继续测试：\n\n");
	int m; i=0,j=2;
	printf("m=i++取值为：%d\n",m=i++); //0
	printf("j++取值为：%d\n\n",j++);  //2
	
	printf("------------------------------\n\n");
	printf("再次测试：\n\n");
	int p; i=0;j=2;
	printf("p=i++ && j++  = %d\n",p=i++ && j++); //0 
	
}

output:

For this question, please combine the following three tests "Eat". I was "going through the cracks" during the testing process. In the end, it can be seen that && is actually used as the separator. First, the value on the left of && is calculated, that is, k=0, i=1, and the value on the left of && is 0, so the right side is no longer calculated (the right side is not executed), so we get The result of i=1, j=2, k=0 is obtained. In fact, the final result of the expression k=i++ && j++ ; is 0. ( 0 &&（）=0)
If the order of operations is wrong, it is easy to make mistakes, just as the "wrong idea" written in the comments.

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