C语言 警告：表达式结果未使用
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warning: expression result unused
提问by user2840926
char change(const char c){
(c >= 'A')&&(c <= 'M') ? (c+'N'-'A') :
((c >= 'N')&&(c <= 'Z') ? (c-('N'-'A')) :
((c >='a')&&(c <= 'm') ? (c+'n'-'a') :
((c >= 'n')&&(c <= 'z') ? (c-('n'-'a')) : c )));
}
Why I get "warning: expression result unused" and "error: control reaches end of non-void function [-Werror,-Wreturn-type]"?
为什么我会收到“警告：未使用的表达式结果”和“错误：控制到达非空函数 [-Werror,-Wreturn-type] 的末尾”？
回答by dasblinkenlight
You get the warning because the expression gets calculated, and then the result is dropped. This is related to the "reaching the end of the function without returning a value" error: adding returnin front of the expression will fix both:
您收到警告是因为计算了表达式，然后删除了结果。这与“到达函数末尾而不返回值”错误有关：return在表达式前面添加将同时修复：
char change(const char c) {
return (c >= 'A') && (c <= 'M') ?
(c+'N'-'A') : ((c >= 'N') && (c <= 'Z') ?
(c-('N'-'A')) : ((c >='a') && (c <= 'm') ?
(c+'n'-'a') : ((c >= 'n') && (c <= 'z') ?
(c-('n'-'a')) : c )));
}
