The shown code snippet, defines a member function named twoSum that has the return type of vector<int> and has 2 parameters. The first parameter named nums is an lvalue reference to a non-const vector<int> while the second parameter named target is an int.
I was wondering what the meaning of the & is
The & in the first parameter nums of the member function means that nums is an lvalue reference to a non-const vector<int>. Meaning, the argument that will be passed to the nums parameter, will be passed by reference instead of passed by value. That is, inside the member function twoSums, the nums refer to the original vector<int> that is passed as an argument.
Also note that there should be a return statement inside the member function since the return type of the member function is non-void.
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显示的代码片段定义了一个名为
twoSum的成员函数，其返回类型为vector并具有2 个参数< /强>。第一个名为nums的参数是对非常量vector的左值引用，而第二个参数名为target code> 是一个int。成员函数的第一个参数
nums中的&表示nums是一个对非常量的左值 代码>向量nums参数的参数将通过引用传递，而不是通过值传递。也就是说，在成员函数twoSums内，nums引用作为参数传递的原始vector。另请注意，成员函数内部应该有 return 语句，因为成员函数的返回类型是非 void。
The shown code snippet, defines a member function named
twoSumthat has the return type ofvector<int>and has 2 parameters. The first parameter namednumsis an lvalue reference to a non-constvector<int>while the second parameter namedtargetis anint.The
&in the first parameternumsof the member function means thatnumsis an lvalue reference to a non-constvector<int>. Meaning, the argument that will be passed to thenumsparameter, will be passed by reference instead of passed by value. That is, inside the member functiontwoSums, thenumsrefer to the originalvector<int>that is passed as an argument.Also note that there should be a return statement inside the member function since the return type of the member function is non-void.