在 C++ 中比较数组的相等性
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Comparing arrays for equality in C++
提问by vladinkoc
Can someone please explain to me why the output from the following code is saying that arrays are not equal?
有人可以向我解释为什么以下代码的输出说数组不相等吗？
int main()
{
int iar1[] = {1,2,3,4,5};
int iar2[] = {1,2,3,4,5};
if (iar1 == iar2)
cout << "Arrays are equal.";
else
cout << "Arrays are not equal.";
return 0;
}
回答by Praetorian
if (iar1 == iar2)
Here iar1and iar2are decayingto pointers to the first elements of the respective arrays. Since they are two distinct arrays, the pointer values are, of course, different and your comparison tests not equal.
这里iar1和iar2被衰减的指针到相应的阵列的第一个元素。由于它们是两个不同的数组，因此指针值当然不同并且您的比较测试不相等。
To do an element-wise comparison, you must either write a loop; or use std::arrayinstead
要进行逐元素比较，您必须编写一个循环；或std::array改用
std::array<int, 5> iar1 {1,2,3,4,5};
std::array<int, 5> iar2 {1,2,3,4,5};
if( iar1 == iar2 ) {
// arrays contents are the same
} else {
// not the same
}
回答by InstallingXubuntuRightNow
Since nobody mentioned it yet, you can compare arrays with the std::equalalgorithm:
由于还没有人提到它，您可以将数组与std::equal算法进行比较：
int iar1[] = {1,2,3,4,5};
int iar2[] = {1,2,3,4,5};
if (std::equal(std::begin(iar1), std::end(iar1), std::begin(iar2)))
cout << "Arrays are equal.";
else
cout << "Arrays are not equal.";
You need to include <algorithm>and <iterator>. If you don't use C++11 yet, you can write:
您需要包括<algorithm>和<iterator>。如果你还没有使用 C++11，你可以这样写：
if (std::equal(iar1, iar1 + sizeof iar1 / sizeof *iar1, iar2))
回答by Fred Larson
You're not comparing the contents of the arrays, you're comparing the addresses of the arrays. Since they're two separate arrays, they have different addresses.
您不是在比较数组的内容，而是在比较数组的地址。由于它们是两个独立的数组，因此它们具有不同的地址。
Avoid this problem by using higher-level containers, such as std::vector, std::deque, or std::array.
通过使用更高级别的容器，如避免这个问题std::vector，std::deque或std::array。
回答by Paul S.
Array is not a primitive type, and the arrays belong to different addressesin the C++ memory.
数组不是原始类型，数组在C++内存中属于不同的地址。
回答by Saurav Sahu
If you are reluctant to change your existing code to std::array, then use a couple of methods instead which takes non-type template arguments:
如果您不愿意将现有代码更改为std::array，则使用一些方法来代替，它们采用非类型模板参数：
//Passed arrays store different data types
template <typename T, typename U, int size1, int size2>
bool equal(T (&arr1)[size1], U (&arr2)[size2] ){
return false;
}
//Passed arrays store SAME data types
template <typename T, int size1, int size2>
bool equal(T (&arr1)[size1], T (&arr2)[size2] ){
if(size1 == size2) {
for(int i = 0 ; i < size1; ++i){
if(arr1[i] != arr2[i]) return false;
}
return true;
}
return false;
}
Here is the demo. Note that, while calling, we just need to pass the array variables e.g. equal(iar1, iar2)in your case, no need to pass the size of arrays.
这是演示。请注意，在调用时，我们只需要传递数组变量，例如equal(iar1, iar2)在您的情况下，不需要传递数组的大小。
回答by Xiangyi Meng
Nobody mentions memcmp? This is also a good choice.
没人提memcmp？这也是一个不错的选择。
/* memcmp example */
#include <stdio.h>
#include <string.h>
int main ()
{
char buffer1[] = "DWgaOtP12df0";
char buffer2[] = "DWGAOTP12DF0";
int n;
n=memcmp ( buffer1, buffer2, sizeof(buffer1) );
if (n>0) printf ("'%s' is greater than '%s'.\n",buffer1,buffer2);
else if (n<0) printf ("'%s' is less than '%s'.\n",buffer1,buffer2);
else printf ("'%s' is the same as '%s'.\n",buffer1,buffer2);
return 0;
}
回答by Vachaspati Mishra
Both store memory addresses to the first elements of two different arrays. These addresses can't be equal hence the output.
两者都将内存地址存储到两个不同数组的第一个元素。这些地址不能相等，因此输出。
回答by Rahul Tripathi
You are comparing the addresses instead of the values.
您正在比较地址而不是值。
回答by vladinkoc
When we use an array, we are really using a pointer to the first element in the array. Hence, this condition if( iar1 == iar2 )actually compares two addresses. Those pointers do not address the same object.
当我们使用数组时，我们实际上是在使用指向数组中第一个元素的指针。因此，此条件if( iar1 == iar2 )实际上是比较两个地址。这些指针不指向同一个对象。
回答by slugonamission
Right. In most, if not all implementations of C, the array identifier can be implicitly casted to a pointer to the first element (i.e. the first element's address). What you're doing here is comparing those addresses, which is obviously wrong.
对。在大多数（如果不是所有的）C 实现中，数组标识符可以隐式转换为指向第一个元素（即第一个元素的地址）的指针。你在这里做的是比较这些地址，这显然是错误的。
Instead, you need to iterate over both arrays, checking each element against each other. If you get to the end of both without a failure, they're equal.
相反，您需要遍历两个数组，相互检查每个元素。如果你在没有失败的情况下完成了两者，那么它们是平等的。
