可能简单但令人困惑的分配错误 [英] Probably simple but confusing assignment error

问题描述

#include <stdio.h>

int main() {
    char gradesList[5];
    gradesList[2] = "X";
    printf("%c", gradesList[2]);
}

当我尝试运行此代码时，出现以下错误:

When I try to run this code I get these errors:

指向整数转换的指针不兼容

Incompatible pointer to integer conversion

赋值从指针进行整数转换而无需强制转换

Assignment makes integer from pointer without a cast

推荐答案

您必须分配char而不是pointer to a string literal.使用'代替"

You have to assign a char not a pointer to a string literal. Use ' instead of "

gradesList[2] = 'X';

---

在C中，字符串文字使用双qoutes表示，即".并且char用单引号表示，即'.

---

In C string literals are represented using double qoutes, i.e. ". And char are represented using single quotes, i.e. '.

由于已将gradesList声明为char数组.并尝试将指针分配给字符串文字，您会收到此错误.

Since you have declared gradesList as a char array. And are trying to assign a pointer to a string literal, you are getting this error.

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