通过＆＃x2018; strcmp＆＃x2019的参数1和2从整数中制作指针，而无需演员[-wint转换]
您好，所以我一直在工作一些prgramme，这是一个计算器（我是初学者），并且在当时的代码结束时可以看到，如果strcmp不起作用，则两个。 VSCODE告诉我（对于StrCMP）发生了例外。分割故障。但是GCC告诉我什么是什么。
#include <stdio.h>
#include <string.h>
int main()
{
float num1;
float num2;
float anwser;
int rnum = 1;
int hi = 0;
char operator;
char ifyorn;
char y = 'y';
char n = 'n';
while (hi == 0)
{
printf("Enter operator +, -, /, x: ");
scanf(" %c", &operator);
printf("Enter num %d :", rnum++);
scanf("%f", &num1);
printf("Enter num %d :", rnum++);
scanf("%f", &num2);
switch (operator)
{
case '+':
anwser = num1 + num2;
printf("Do you want to continue y/n \n");
scanf(" %c", &ifyorn);
break;
case '-':
anwser = num1 - num2;
printf("Do you want to continue y/n \n");
scanf(" %c", &ifyorn);
break;
case 'x':
anwser = num1 * num2;
printf("Do you want to continue y/n \n");
scanf(" %c", &ifyorn);
break;
case '/':
anwser = num1 / num2;
printf("Do you want to continue y/n \n");
scanf(" %c", &ifyorn);
break;
default:
printf("This is not a valid character please try again :(");
break;
}
if(strcmp (ifyorn, n) == 0)
{
printf("%f", anwser);
hi == 1;
}
if(strcmp (ifyorn, y) == 0)
{
hi == 0;
}
}
}
Hello so I've been working a little prgramme which is sort of a calculator (I'm a beginner) and well as you can see in the tittle at then end of the code, the two if strcmp doesn't work. And vscode is telling me (for the strcmp) Exception has occurred. Segmentation fault. But gcc is telling me what is in the tittle.
#include <stdio.h>
#include <string.h>
int main()
{
float num1;
float num2;
float anwser;
int rnum = 1;
int hi = 0;
char operator;
char ifyorn;
char y = 'y';
char n = 'n';
while (hi == 0)
{
printf("Enter operator +, -, /, x: ");
scanf(" %c", &operator);
printf("Enter num %d :", rnum++);
scanf("%f", &num1);
printf("Enter num %d :", rnum++);
scanf("%f", &num2);
switch (operator)
{
case '+':
anwser = num1 + num2;
printf("Do you want to continue y/n \n");
scanf(" %c", &ifyorn);
break;
case '-':
anwser = num1 - num2;
printf("Do you want to continue y/n \n");
scanf(" %c", &ifyorn);
break;
case 'x':
anwser = num1 * num2;
printf("Do you want to continue y/n \n");
scanf(" %c", &ifyorn);
break;
case '/':
anwser = num1 / num2;
printf("Do you want to continue y/n \n");
scanf(" %c", &ifyorn);
break;
default:
printf("This is not a valid character please try again :(");
break;
}
if(strcmp (ifyorn, n) == 0)
{
printf("%f", anwser);
hi == 1;
}
if(strcmp (ifyorn, y) == 0)
{
hi == 0;
}
}
}
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评论（2）
变量
ifyorn，y和n被声明具有类型char。函数
strcmp期望指示指向字符串的指针类型char *的参数。因此，如果陈述
且不
正确。取而代之的是，您应该编写
，也
应该写这些语句中的比较操作员，而不是分配
，
您需要编写
增加
变量
rnum看起来毫无和
意义
默认您应该添加更多语句The variables
ifyorn,yandnare declared having the typechar.The function
strcmpexpects arguments of the pointer typechar *that point to strings.So these if statements
and
are incorrect. Instead you should write
and
Also instead of assignments you are using the comparison operator in these statements
and
You need to write
and
Increasing the variable
rnumlooks senselessWhy not just to write
And in the code snippet under the label
defaultyou should add one more statement您不必对这个家伙刻薄，他正在学习。
您会遇到此错误，因为您将字符传递到strcmp（）而不是指针到字符。
这是有关该功能的更多信息。
You don't have to be mean to the guy ,he is learning.
You are getting this error because you are passing characters to strcmp() instead of pointers to characters.
Here is more information regarding that function.