C ++错误-表达式必须具有整数或枚举类型-从具有串联的字符串中获取此类型? [英] C++ error -- expression must have integral or enum type -- getting this from a string with concatenation?

问题描述

C ++错误表达式必须具有整数或枚举类型，该字符串是从带有串联的字符串中获取的吗?

C++ error expression must have integral or enum type getting this from a string with concatenation?

因此在C ++中某个类的 toString()中，我有以下代码:

So in the toString() of a class in C++ I have the code:

string bags = "Check in " + getBags() + " bags";

我以为可以声明这样的字符串?(我来自Java背景，试图学习C ++).但是，在Visual Studio中对 bags 下划线了，问题是:

I thought I could declare a string like this? (I'm coming from a Java background and trying to learn C++). The bags is underlined in Visual Studio though and the problem is:

表达式必须具有整数或枚举类型.

expression must have integral or enum type.

getBags()仅返回 int .

发生这种情况的另一个示例是:

Another example where this happens is with:

string totalPrice = "Grand Total: " + getTotalPrice();

getTotalPrice()返回 float ，并且带有错误的下划线是.

getTotalPrice() returns a float and is what is underlined with the error.

但是如果我把这样的行放进去:

But then if I put in a line like:

string blah = getBags() + "blah";

没有错误.

我在这里不明白什么?

推荐答案

签入" 实际上是 const char * .向其添加 getBags()(一种 int )会生成另一个 const char * .生成编译器错误是因为无法添加两个指针.

"Check in " is actually a const char *. Adding getBags() (an int) to it yields another const char*. The compiler error is generated because you cannot add two pointers.

在连接它们之前，您需要将签入" 和 getBags()都转换为字符串:

You need to convert both "Check in " and getBags() to strings before concatenating them:

string bags = std::string("Check in ") + std::to_string(getBags()) + " bags";

袋子" 将隐式转换为 string .

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