请帮忙什么是逻辑错误...编译并运行代码并检查.. Y是否会出现意外错误 [英] Please help whats is the logical error ... Compile and run the code and check .. Y does it give unexpected error

问题描述

的

＃包括< stdio.h中> 
 #include< conio.h> 
 #include< stdlib.h> 
 #include< math.h> 
 
 int main（）
 {
 int a = 0，b = 0; 
 int sum，diff，mult，R; 
 float div; 
 char c，ch; 
 
 do 
 {
 printf（\ nn \ nn\\\\\\\\\\\\\\\\\\\\\\\ 
 scanf（％d，& a）; //取一个例如7 
 printf的值（\ n \\\\\\\\\\\\\\\\\\\\\\\\ 
 scanf（％d，& b）; //取b的值，例如4 
 printf（\\\\ _______________________________）; 
 printf（\t |输入所需的操作员|）; 
 printf（\t _______________________________）; 
 printf（\\ \\ n \ n \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\新闻* \ n \ n \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\
\\\
）; 
 scanf（％s，& c）; 
 printf（\\\\\ n要执行的操作是'％c'按任何键继续，c）; 
 getch（）; 
 
 if（c =='+'|| c ==' - '|| c =='*'|| c =='/'|| c =='％'|| c =='0'）
 {
 printf（\ nn\ n | RESULTS |）; 
 
 if（c ==' - '）
 {
 diff = a-b; 
 printf（\ n \ nDDFERFER OF％d AND％d IS：％d \ n \ n，a，b，diff）; //这里它应该输出为7-4 = 3但它给出了7-0 = 7。为什么这么
 getch（）; 
}否则if（c =='+'）
 {
 sum = a + b; 
 printf（\ n \\ n \\ n \\ n \\ n \\ n \\ n分配％d和％d IS：％d \ n \\ n，a，b，sum）; 
 getch（）; 
}否则if（c =='*'）
 {
 mult = a * b; 
 printf（\ n \ n \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ 
 getch（）; 
}否则if（c =='/'）
 {
 div = a /（float）b; 
 printf（\ n \ n \\\\\\\,％d和％d IS：％f \ n \ n，a，b，div）; 
 getch（）; 
}否则if（c =='％'）
 {
 R = a％b; 
 printf（\ n \\\\\\\\\\\\\\\\\\\\\\\\ 
 getch（）; 
} 
} 
 else {
 printf（\ n \ nn \ n \ tt \ t \ t无效运营商输入正确的运营商）; 
 getch（）; 
} 
 
} while（c！='0'）; 
返回0; 
} 

我的尝试：



i尝试在begning初始化值，即int a = 7，b = 4;

通过这样做它给出正确的东西但是当我们使用scanf从用户获取输入时它不会作品

解决方案

仔细阅读 scanf文档 。另请查看示例代码以了解字符串的使用。



您的c是字符而不是字符串。



所以正确的代码是：

 scanf（ ％C，和C）; 

#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
#include<math.h>

int main()
{
	int a=0,b=0;
	int sum,diff,mult,R;
	float div;
	char c,ch;

do
{
	printf("\n\n\tENTER THE FIRST operand   ");
	scanf("%d",&a); //takes value for a for eg 7
	printf("\n\n\tENTER THE SECOND operand   ");
	scanf("%d",&b);//takes value for b for eg 4
	printf("\t\n\n_______________________________");
	printf("\t | ENTER THE DESIRED OPERATOR |");
	printf("\t_______________________________");
	printf("\n\n \t\tFOR SUMMATION PRESS +  \n\n \t\tFOR DIFFERENCE PRESS -  \n\n \t\tFOR PRODUCT PRESS * \n\n \t\tFOR DIVISION PRESS / \n\n \t\tTO FIND REMAINDER ONLY PRESS %% \n\n \t\tPRESS 0 TO EXIT\n\n");
	scanf("%s",&c);
	printf("\t\n\n  OPERATION TO BE PERFORMED IS '%c' PRESS ANY KEY TO CONTINUE",c);
	getch();

 if(c=='+'||c=='-'||c=='*'||c=='/'||c=='%'||c=='0')
{
	printf("\n\n |RESULTS|");

	if(c=='-')
	{	
	diff=a-b;
	printf("\n\nDIFFERENCE OF %d AND %d IS : %d \n\n",a,b,diff);// here it should give output as 7-4=3 but it gives 7-0=7 . why is it so
getch();
}  else if(c=='+')
{	
	sum=a+b;
	printf("\n\n\tSUMMATION OF %d AND %d IS : %d \n\n",a,b,sum);
	getch();
} else if(c=='*')
{	
	mult=a*b;
	printf("\n\n\tPORDUCT OF %d AND %d IS : %d \n\n",a,b,mult);
	getch();
} else if(c=='/')
{	
	div=a/(float)b;
	printf("\n\n\tDIVISION OF %d AND %d IS : %f \n\n",a,b,div);
	getch();
} else if(c=='%')
{	
	R=a%b;
	printf("\n\n\tREMAINDER WHEN %d IS DIVIDED BY %d IS : %d \n\n",a,b,R);
	getch();
}
}
else {
	printf("\n\n\n\t\t\t  INVALID OPERATOR ENTER CORRECT OPERATOR ");
	getch();
}

}while(c!= '0');
return 0;
}

What I have tried:

i tried to initialise values at the begning i.e int a=7,b=4;
by doing this it gives correct thing but when we use scanf for getting input from user it doesnt works

解决方案

Read the scanf documentationcarefully. Take also a look on the example code to understand the string use.

Your c is a char not a string.

So correct code would be:

scanf("%c",&c);

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