警告：控制到达非空函数的结尾（C++）
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warning: control reaches end of non-void function (c++)
提问by Mesmer Lucas
i get this error and can't fix, i'm noob yet,if someone can help me i'll thank you this code come from xmplayer of libxenon (for jtag xbox)
我收到这个错误并且无法修复，我还是菜鸟，如果有人可以帮助我，我会感谢你这个代码来自 libxenon 的 xmplayer（对于 jtag xbox）
(i try search similar error, but i can't find what's wrong)
（我尝试搜索类似的错误，但我找不到问题所在）
int FileSortCallback(const void *f1, const void *f2) {
/* Special case for implicit directories */
if (((BROWSERENTRY *) f1)->filename[0] == '.' || ((BROWSERENTRY *) f2)->filename[0] == '.') {
if (strcmp(((BROWSERENTRY *) f1)->filename, ".") == 0) {
return -1;
}
if (strcmp(((BROWSERENTRY *) f2)->filename, ".") == 0) {
return 1;
}
if (strcmp(((BROWSERENTRY *) f1)->filename, "..") == 0) {
return -1;
}
if (strcmp(((BROWSERENTRY *) f2)->filename, "..") == 0) {
return 1;
}
}
/* If one is a file and one is a directory the directory is first. */
if (((BROWSERENTRY *) f1)->isdir && !(((BROWSERENTRY *) f2)->isdir)) return -1;
if (!(((BROWSERENTRY *) f1)->isdir) && ((BROWSERENTRY *) f2)->isdir) return 1;
//Ascending Name
if (XMPlayerCfg.sort_order == 0) {
return stricmp(((BROWSERENTRY *) f1)->filename, ((BROWSERENTRY *) f2)->filename);
}
//Descending Name
else if (XMPlayerCfg.sort_order == 1) {
return stricmp(((BROWSERENTRY *) f2)->filename, ((BROWSERENTRY *) f1)->filename);
}
//Date Ascending
else if (XMPlayerCfg.sort_order == 2) {
if ( ((BROWSERENTRY *) f2)->date == ((BROWSERENTRY *) f1)->date) { //if date is the same order by filename
return stricmp(((BROWSERENTRY *) f2)->filename, ((BROWSERENTRY *) f1)->filename);
} else {
return ((BROWSERENTRY *) f1)->date - ((BROWSERENTRY *) f2)->date;
}
}
//Date Descending
else if (XMPlayerCfg.sort_order == 3) {
if ( ((BROWSERENTRY *) f2)->date == ((BROWSERENTRY *) f1)->date) { //if date is the same order by filename
return stricmp(((BROWSERENTRY *) f1)->filename, ((BROWSERENTRY *) f2)->filename);
} else {
return ((BROWSERENTRY *) f2)->date - ((BROWSERENTRY *) f1)->date;
}
}
}
回答by dasblinkenlight
The compiler analyzes your code, and sees that a return statement will be executed for all values of sort_orderbetween 0and 5, inclusive. However, if the sort_orderis negative or more than 5, the code would reach the end of the function without a return statement; that is why the compiler issues a warning.
编译器会分析您的代码，并发现将对sort_order介于0和之间的所有值执行 return 语句5，包括。但是，如果sort_order为负数或大于5，则代码将在没有返回语句的情况下到达函数的末尾；这就是编译器发出警告的原因。
Note that it may not be possible for sort_orderto be set to a negative number or a number over 5because of constraints in other parts of your code. However, the compiler does not know any of that, so it thinks that sort_ordercould have any value.
请注意，由于代码其他部分的限制，可能无法将sort_order其设置为负数或数字5。但是，编译器不知道任何这些，所以它认为这sort_order可能有任何价值。
To fix this problem, add an unconditional return statement at the end.
要解决此问题，请在末尾添加无条件 return 语句。
