C++ 错误：[函数名] 之前的预期初始化程序
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C ++ error: a expected initializer before [function name]
提问by WorkerBee
I am refreshing my self on C++ (have not did it since school) and I wrote a simple program just to mess around. My problem is when I compile the program it chokes stating "error: expected initializer before 'stringThing'" is there a reason why this is doing this? I know this may be a noob question so I checked stackoverflow and could not find any relevant questions that gave me a answer.
我正在用 C++ 更新我的自我（从学校开始就没有这样做过），我写了一个简单的程序只是为了捣乱。我的问题是，当我编译程序时，它会提示“错误：'stringThing' 之前的预期初始化程序”是否有这样做的原因？我知道这可能是一个菜鸟问题，所以我检查了 stackoverflow 并找不到任何给我答案的相关问题。
*I am using GNU GCC compiler
*我正在使用 GNU GCC 编译器
Code:
代码：
#include <iostream>
using namespace std;
void string stringThing (string shiftdir, string &teststring)
{
if (shiftdir == "right")
{
teststring = teststring >> " " >> "Bit Shifted right";
}
else
{
teststring = teststring << " " << "Bit Shifted left";
}
}
int main()
{
string test;
cout << stringThing("right", "I have done a ") << endl;
return 0;
}
采纳答案by Olaf Dietsche
The return type for stringThingmust be either voidorstring, not both. You also must include <string>, if you want to use string.
的返回类型stringThing必须是void或string，而不是两者。<string>如果要使用字符串，还必须包含。
Since you want to output the return value of stringThing()in main, I guess it should be
既然要输出stringThing()in的返回值main，我想应该是
std::string stringThing (std::string shiftdir, const std::string &teststring)
But then, you must also return a string from your function
但是，您还必须从函数中返回一个字符串
if (shiftdir == "right")
return teststring + " " + "Bit Shifted right";
else
return teststring + " " + "Bit Shifted left";
for example.
例如。
Your parameter std::string &teststringwon't work with your const char*argument. So either declare it as a copy by value stringonly, or better const string&.
您的参数std::string &teststring不适用于您的const char*参数。因此，要么string仅按值将其声明为副本，要么更好const string&。
回答by Lightness Races in Orbit
Return type is … funky
返回类型是……时髦
What is:
什么是：
void string stringThing (string shiftdir, string &teststring)
?
?
Get rid of the first string. Your function returns nothing.
摆脱第一个string。您的函数不返回任何内容。
So, simply:
所以，简单地说：
void stringThing(string shiftdir, string &teststring)
Inclusion missing
包含缺失
You will also need to #include <string>— in some scenarios you may get "lucky" and have it implicitly included by <iostream>, but don't rely on it.
您还需要#include <string>- 在某些情况下，您可能会“幸运”并将其隐式包含在 中<iostream>，但不要依赖它。
