[C Programming Tutorial Experiment] Basic Exercises (3)

content

Code 1: The priority problem

//下列各个表达式取值为？ 
#include <stdio.h>
int main()
{
    
    
	printf("已知: a=3,b=4,c=5\n\n");
	printf("则有：\n\n");
	
	int a=3,b=4,c=5;//显然b==c为假，故取值为0 
	printf("(a+b>c && b==c) = %d\n\n",a+b>c && b==c);
	
	a=3,b=4,c=5;//先看&&右边，b-c=-1（非0）为真，右边取1。而&&左边有个||，又a=3(非0)，故&&左边亦取1。因此整个表达式为1。 
	printf("(a||b+c && b-c) = %d\n\n",a||b+c && b-c);
	
	a=3,b=4,c=5;//!c取0，而无论前边怎么取，取0还是取1，后边有个||，还有1，故最终结果为1 
	printf("(!(a>b)&&!c||1) = %d\n\n",!(a>b)&&!c||1);
 
 	a=3,b=4,c=5;//&&右边，c/2为2,2+4=6（非0），故&&右边取1。再看&&左边，a+b=7,!(a+b)取0，0+c-1=4为非0，故&&左边取1。所以整个表达式取1 
 	printf("(!(a+b)+c-1 && b+c/2) = %d\n",!(a+b)+c-1 && b+c/2);	
 }

output:

The topic is simple, I think I made it clear in the comments!
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Code 2: Accuracy problem

//精度问题。。。 
#include <stdio.h>
int main()
{
    
    
	int x;
	float y=5.5; //y*3=16.500000 
	x=(float)(y*3+((int)y)%4); //首先 (int)y=5,其次 ((int)y)%4=1 
	printf("x=%d\n\n",x);
	
	printf("-----------\n\n");
	printf("测试：\n\n");
	printf("(int)y = %d\n\n",(int)y);
	printf("((int)y) %% 4 = %d\n\n",((int)y)%4);
	printf("y*3=%f\n\n",y*3);
	printf("(float)(y*3+(((int)y)%4) = %f\n",(float)(y*3+((int)y)%4));
	//又由于x是整型，所以17.500000取整为17 
 }

output:

Here, I am a little bit tricky about the printout format of %.

This topic is very basic and simple, and it should be well understood by combining the annotation and testing parts.
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Code three: (?:) expression evaluation

//?:表达式求值 
#include <stdio.h>
int main()
{
    
    
	int x=1,y=1;
	
	int a=1,b=4,c=3,d=2;
	
	printf("!x || y-- = %d\n\n",!x || y--);//!x取值为0,y--取值为1,故0||1取值为1。 
	
	printf("a<b?a:c<d?c:d = %d\n\n",a<b?a:c<d?c:d);
	
	printf("-------------------------\n\n");
	printf("测试：\n\n");
	printf("((a<b)?(a):(c<d?c:d)) = %d\n\n",((a<b)?(a):(c<d?c:d))); 
	//对于上式，a<b为真,直接取a=1,最后结果亦为1 (后边不再执行)。c<d为假，取d=2。 
	
	printf("-------------------------\n\n");
	printf("再次测试:\n\n");
	printf("a<b?a:666 = %d\n\n",a<b?a:666); //1<4 √√ 故取a=1 
	printf("c<d?c:d = %d\n\n",c<d?c:d);  //3<2 ×××故取d=2 
	return 0;
}

Test output:

I think the comment has already said it more clearly!

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Code 4: Relax for a moment ~ ~!

//轻松一刻 ~ ~！下列程序的输出结果为
#include <stdio.h>
int main()
{
    
    
	int a=1,b=2,c;
	int x=97; 
	
	c=1.0/b*a; //首先1.0/b 即 1.0/2 = 0.500000,而后 0.5*a 即0.500000*1=0.500000
	//又因为c为整型，故c最后取整为  0 
	printf("c=%d\n",c);
	
	printf("x=%c\n",x);
	printf("x=%d\n",x);
 }

output:
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