发现它没有给出输出 [英] Findd y it doesnt give output

问题描述

＃包括< stdio.h中> 
 #include< conio.h> 
 #include< stdlib.h> 
 #include< math.h> 
 
 main（）
 {
 
 int a = 0，b = 0; 
 int sum; 
 char c; 
 
 
 
 
 printf（\ n \\\ nnn，第一个操作数）; 
 scanf（％d，& a）; 
 printf（\ n \\\\\\\\\\\\\\\\\\\\ 
 scanf（％d，& b）; 
 
 printf（\ nn \ n \\\\\\\\\\\\\\\\\\\\\\\\ 
 
 scanf（％s，& c）; 
 
 if（c =='+'）
 {
 
 
 printf（\ nnanth | RESULTS |）; 
 
 
 
 
 sum = a + b; 
 printf（\ n \\ n \\ n \\ n \\ n \\ n \\ n分配％d和％d IS：％d \ n \\ n，a，b，sum）; 
 
} 
 else {
 
 printf（\ n \ nn \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ ）; 
 
 
 
} 
} 

我是什么已经尝试过：



试图通过itlf将第二个变量放到0

解决方案

< blockquote>可能是这导致了这个问题：

 char c; 
 ... 
 scanf（％s，& c）; 

当您使用格式％s调用scanf时，它期望 - 并且取出 - 一个以空字符结尾的字符串，而不是单个字符。因为你只为字符串分配一个字符，它会溢出你给它的空间，实际上接下来发生的事情是在编译器编写器的一圈 - 这是一个非法的操作，因此没有定义效果。 />


将c的定义更改为字符数组：

  char   operator  [ 100 ]; 

并读取一个字符串来自用户：

 scanf（ ％s ， operator ）; 

然后检查用户输入的第一个字符：

  if （ operator  [ 0 ] == '  +'）

它应该更好。

％s格式说明符带有 char 变量是一个错误。

试试，例如

  #include   <   stdio.h  >  
   #include   <   stdlib.h  >  
  
  int  main（）
 {
  int  a =  0 ，b =  0  ; 
  int  sum; 
 
 printf（  \ nn \ nn\\\\\\\\\\\\\\\跨度>）; 
 scanf（ ％d，& a）; 
 printf（  \ n \ nn \ tnTERTER SECOND operand）; 
 scanf（ ％d，& b）; 
 
 printf（  \ n \ n \\\\\\\\\\\\\\\\\\\\\\ ）; 
 
  while （getchar（）！= '  +'）
 {
 printf（  \ n无效操作员输入正确的操作员）; \\ n \\ n \\ n \\ n \\ n \\\\\ 
} 
 printf（  \ n\ n | RESULTS | ）; 
 sum = a + b; 
 printf（  \ n \ n \\\\\ \\\
\\\
，A，b，和）; 
  return   0 ; 
} 

您已在请帮忙什么是逻辑错误...编译并运行代码并检查.. Y是否会产生意外错误 [ ^ ]，并收到一些建议。请不要重新发布相同的问题。

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

 main()
{

	int a=0,b=0;
	int sum;
	char c;


	

	printf("\n\n\tENTER THE FIRST operand   ");
	scanf("%d",&a);
	printf("\n\n\tENTER THE SECOND operand   ");
	scanf("%d",&b);
	
	printf("\n\n \t\tFOR SUMMATION PRESS +  ");

	scanf("%s",&c);
	
 if(c=='+')
{


	printf("\n\n |RESULTS|");

	
	
	
	sum=a+b;
	printf("\n\n\tSUMMATION OF %d AND %d IS : %d \n\n",a,b,sum);

}
else {

	printf("\n\n\n\t\t\t  INVALID OPERATOR ENTER CORRECT OPERATOR ");



}
}

What I have tried:

tried to assighn values bt puts second variable to 0 by itslf

解决方案

Probably it's this that causes the problem:

char c;
...
scanf("%s",&c);

When you call scanf with a format of "%s", it expects - and fetches - a null terminated string, not a single character. Since you only allocate a single character for the string, it will overflow the space you gave it, and in practice what happens next is in the lap of the compiler writer - it's an illegal operation and so the effects aren't defined.

Change the definition of c to an array of characters:

char operator[100];

And read a string from the user:

scanf("%s",operator);

Then check the first character the user typed:

if (operator[0] == '+')

It should work better.

Quote:

scanf("%s",&c);

using the "%s" format specifier with a char variable is an error.
Try, for instance

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

int  main()
{
  int a=0,b=0;
  int sum;

  printf("\n\n\tENTER THE FIRST operand   ");
  scanf("%d", &a);
  printf("\n\n\tENTER THE SECOND operand   ");
  scanf("%d",&b);

  printf("\n\n \t\tFOR SUMMATION PRESS +  ");

  while ( getchar() != '+')
  {
    printf("\n\n\n\t\t\t  INVALID OPERATOR ENTER CORRECT OPERATOR ");
  }
  printf("\n\n |RESULTS|");
  sum=a+b;
  printf("\n\n\tSUMMATION OF %d AND %d IS : %d \n\n",a,b,sum);
  return 0;
}

You already posted this at Please help whats is the logical error ... Compile and run the code and check .. Y does it give unexpected error[^], and received some suggestions. Please do not repost the same question.

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