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c - 为什么将我的输入分配给枚举会导致段错误?

转载 作者:行者123 更新时间:2023-11-30 18:22:13 25 4
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我有以下程序

#include <stdio.h>
#include "dbg.h"

#define MAX_DATA 100

typedef enum EyeColor{
BLUE_EYES, GREEN_EYES, BROWN_EYES, BLACK_EYES, OTHER_EYES
} EyeColor;

const char *EYE_COLOR_NAMES[] = {
"Blue", "Green", "Brown", "Black", "Other"
};

typedef struct Person {
int age;
char first_name[MAX_DATA];
char last_name[MAX_DATA];
EyeColor eyes;
float income;
} Person;

int main(int argc, char *argv[]){
Person you = {.age = 0};
int i = 0;
char *in = NULL;

printf("What's your First Name? ");
in = fgets(you.first_name, MAX_DATA-1, stdin); //fgets is much better
check(in != NULL, "Failed to read first name.");

printf("What's your Last Name? ");
in = fgets(you.last_name, MAX_DATA-1, stdin);
check(in != NULL, "Failed to read last name.");

printf("How old are you? ");
int rc = fscanf(stdin, "%d", &you.age);
check(rc > 0, "You have to enter a number.");

printf("What color are your eyes:\n");
for(i = 0; i <= OTHER_EYES; i++) {
printf("%d) %s\n", i+1, EYE_COLOR_NAMES[i]);
}
printf("> ");

int eyes = -1;
rc = fscanf(stdin, "%d", &eyes);
check(rc > 0, "You have to enter a number.");

you.eyes = eyes - 1;
//check(you.eyes <= OTHER_EYES && you.eyes >= 0, "Do it right, that's not an option."); // if you dont check this, seg fault can happen because of ...

printf("How much do you make an hour? ");
rc = fscanf(stdin, "%f", &you.income);
check(rc > 0, "Enter a floating point number.");

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

printf("First Name: %s", you.first_name);
printf("Last Name: %s", you.last_name);
printf("Age: %d\n", you.age);
printf("Eyes: %s\n", EYE_COLOR_NAMES[you.eyes]);
printf("Income: %f\n", you.income);

return 0;
error:
return 1;
}

如果我跑:$./ex24

我给出以下输入:

What's your First Name? a
What's your Last Name? b
How old are you? 1
What color are your eyes:
1) Blue
2) Green
3) Brown
4) Black
5) Other
> 1000
How much do you make an hour? 1
------ RESULTS ------
First Name: a
Last Name: b
Age: 1
Segmentation fault (core dumped)

这个段错误是如何发生的? enum 不是在我的内存中分配 int 类型,应该可以容纳 1000 吗?

编辑1:将一些文本格式化为代码。编辑2:注释掉检查。

最佳答案

枚举是否可以达到 1000 并不重要;您的数组没有 1000 个元素。
没有可打印的第 1000th 眼睛颜色。

这是您应该构建的测试用例:

#include <iostream>

const char* EYE_COLOR_NAMES[] = { "a", "b", "c", "d", "e" };
const int eyes = 1000;

int main()
{
std::cout << EYE_COLOR_NAMES[eyes] << '\n';
}

我想说,很明显这是行不通的。

您已经在代码中对 you.eyes 进行了范围检查,并且您自己发现只有在进行范围检查时才会发生段错误。因此,只需保留范围检查即可。

关于c - 为什么将我的输入分配给枚举会导致段错误?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27627333/

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