错误:将“const char[5]"分配给“char[10]"时的类型不兼容

2022-01-12 00:00:00 char c++

我已将 c 定义为

char c[][10]

在函数定义中并像 c[i]="gray";

怎么了?我在网上搜索，它显示相同的语法.

Whats wrong? I searched on net, it shows the same syntax.

谢谢.

推荐答案

您不能对数组使用赋值 (=).如果您将 c 更改为指针数组，这可能会起作用，具体取决于您需要使用它做什么.

You cannot use assignment (=) on an array. If you change c to an array of pointers, that might work, depending on what you need to do with it.

const char *c[20]; c[i] = "gray";

或者如果声明的类型必须是数组数组，你可以使用 strncpy:

Or if the declared type must be array of arrays, you could use strncpy:

char c[20][10]; strncpy(c[i], "gray", sizeof(c[i]));

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