C++ 将结构传递给函数
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Passing structs to functions
提问by Phorce
I am having trouble understanding how to pass in a struct (by reference) to a function so that the struct's member functions can be populated. So far I have written:
我无法理解如何将结构（通过引用）传递给函数，以便可以填充结构的成员函数。到目前为止，我已经写了：
bool data(struct *sampleData)
{
}
int main(int argc, char *argv[]) {
struct sampleData {
int N;
int M;
string sample_name;
string speaker;
};
data(sampleData);
}
The error I get is:
我得到的错误是：
C++ requires a type specifier for all declarations bool data(const &testStruct)
C++ 要求所有声明的类型说明符 bool data(const &testStruct)
I have tried some examples explained here: Simple way to pass temporary struct by value in C++?
我已经尝试过这里解释的一些示例：Simple way to pass临时结构 by value in C++?
Hope someone can Help me.
希望可以有人帮帮我。
回答by Nikos C.
First, the signature of your data() function:
首先，你的 data() 函数的签名：
bool data(struct *sampleData)
cannot possibly work, because the argument lacks a name. When you declare a function argument that you intend to actually access, it needs a name. So change it to something like:
不可能工作，因为参数缺少名称。当您声明要实际访问的函数参数时，它需要一个名称。所以把它改成这样：
bool data(struct sampleData *samples)
But in C++, you don't need to use structat all actually. So this can simply become:
但是在 C++ 中，您struct实际上根本不需要使用。所以这可以简单地变成：
bool data(sampleData *samples)
Second, the sampleDatastruct is not known to data() at that point. So you should declare it before that:
其次，此时sampleDatadata() 不知道该结构体。所以你应该在此之前声明它：
struct sampleData {
int N;
int M;
string sample_name;
string speaker;
};
bool data(sampleData *samples)
{
samples->N = 10;
samples->M = 20;
// etc.
}
And finally, you need to create a variable of type sampleData. For example, in your main() function:
最后，您需要创建一个类型为 的变量sampleData。例如，在您的 main() 函数中：
int main(int argc, char *argv[]) {
sampleData samples;
data(&samples);
}
Note that you need to pass the address of the variable to the data() function, since it accepts a pointer.
请注意，您需要将变量的地址传递给 data() 函数，因为它接受一个指针。
However, note that in C++ you can directly pass arguments by reference and don't need to "emulate" it with pointers. You can do this instead:
但是，请注意，在 C++ 中，您可以直接通过引用传递参数，而无需使用指针“模拟”它。你可以这样做：
// Note that the argument is taken by reference (the "&" in front
// of the argument name.)
bool data(sampleData &samples)
{
samples.N = 10;
samples.M = 20;
// etc.
}
int main(int argc, char *argv[]) {
sampleData samples;
// No need to pass a pointer here, since data() takes the
// passed argument by reference.
data(samples);
}
回答by Inisheer
bool data(sampleData *data)
{
}
You need to tell the method which type of struct you are using. In this case, sampleData.
您需要告诉方法您正在使用哪种类型的结构。在这种情况下，示例数据。
Note: In this case, you will need to define the struct prior to the method for it to be recognized.
注意：在这种情况下，您需要在要识别的方法之前定义结构。
Example:
例子：
struct sampleData
{
int N;
int M;
// ...
};
bool data(struct *sampleData)
{
}
int main(int argc, char *argv[]) {
sampleData sd;
data(&sd);
}
Note 2: I'm a C guy. There may be a more c++ish way to do this.
注2：我是一个C家伙。可能有一种更 c++ 的方式来做到这一点。
回答by Hossein Jandaghi
Passing structs to functions by reference: simply :)
通过引用将结构传递给函数：简单地:)
#define maxn 1000
struct solotion
{
int sol[maxn];
int arry_h[maxn];
int cat[maxn];
int scor[maxn];
};
void inser(solotion &come){
come.sol[0]=2;
}
void initial(solotion &come){
for(int i=0;i<maxn;i++)
come.sol[i]=0;
}
int main()
{
solotion sol1;
inser(sol1);
solotion sol2;
initial(sol2);
}
