编译器错误 C3538

在声明符列表中，“auto”必须始终推导为同一类型

声明列表中的所有已声明变量不解析为同一类型。

更正此错误

1. 确保列表中的所有 auto 声明都推导为同一类型。

示例

下面的语句生成 C3538。 每条语句声明多个变量，但 auto 关键字的每次使用都不推导为同一类型。

// C3538.cpp
// Compile with /Zc:auto
// C3538 expected
int main()
{
// Variable x1 is a pointer to char, but y1 is a double.
   auto * x1 = "a", y1 = 3.14;
// Variable c is a char and c1 is char*, but c2, and c3 are pointers to pointers.
   auto c = 'a', *c1 = &c, * c2 = &c1, * c3 = &c2;
// Variable x2 is an int, but y2 is a double and z is a char.
   auto x2(1), y2(0.0), z = 'a';
// Variable a is a pointer to int, but b is a pointer to double.
   auto *a = new auto(1), *b = new auto(2.0);
   return 0;
}

另请参阅

auto 关键字