C++ 将两个 int 数组连接成一个更大的数组
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C++ concatenate two int arrays into one larger array
提问by imkendal
Is there a way to take two int arrays in C++
有没有办法在 C++ 中获取两个 int 数组
int * arr1;
int * arr2;
//pretend that in the lines below, we fill these two arrays with different
//int values
and then combine them into one larger array that contains both arrays' values?
然后将它们组合成一个包含两个数组值的更大数组？
回答by texasbruce
int * result = new int[size1 + size2];
std::copy(arr1, arr1 + size1, result);
std::copy(arr2, arr2 + size2, result + size1);
Just suggestion, vectorwill do better as a dynamic array rather than pointer
只是建议，vector作为动态数组而不是指针会做得更好
回答by Reed Copsey
If you're using arrays, you need to allocate a new array large enough to store all of the values, then copy the values into the arrays. This would require knowing the array sizes, etc.
如果您使用数组，则需要分配一个足够大的新数组来存储所有值，然后将这些值复制到数组中。这需要知道数组大小等。
If you use std::vectorinstead of arrays (which has other benefits), this becomes simpler:
如果你使用std::vector而不是数组（它有其他好处），这会变得更简单：
std::vector<int> results;
results.reserve(arr1.size() + arr2.size());
results.insert(results.end(), arr1.begin(), arr1.end());
results.insert(results.end(), arr2.begin(), arr2.end());
回答by user1095108
Another alternative is to use expression templates and pretend the two are concatenated (lazy evaluation). Some links (you can do additional googling):
另一种选择是使用表达式模板并假装两者连接在一起（惰性求值）。一些链接（你可以做额外的谷歌搜索）：
http://www10.informatik.uni-erlangen.de/~pflaum/pflaum/ProSeminar/http://www.altdevblogaday.com/2012/01/23/abusing-c-with-expression-templates/ http://aszt.inf.elte.hu/~gsd/halado_cpp/ch06s06.html
http://www10.informatik.uni-erlangen.de/~pflaum/pflaum/ProSeminar/http://www.altdevblogaday.com/2012/01/23/abusing-c-with-expression-templates/ http:// /aszt.inf.elte.hu/~gsd/halado_cpp/ch06s06.html
If you are looking for ease of use, try:
如果您正在寻找易用性，请尝试：
#include <iostream>
#include <string>
int main()
{
int arr1[] = {1, 2, 3};
int arr2[] = {3, 4, 6};
std::basic_string<int> s1(arr1, 3);
std::basic_string<int> s2(arr2, 3);
std::basic_string<int> concat(s1 + s2);
for (std::basic_string<int>::const_iterator i(concat.begin());
i != concat.end();
++i)
{
std::cout << *i << " ";
}
std::cout << std::endl;
return 0;
}
回答by user1095108
Here is the solution for the same-
这是相同的解决方案-
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
void Concatenate(char *s1,char *s2)
{
char s[200];
int len1=strlen(s1);
int len2=strlen(s2);
int j;
///Define k to store the values on Kth address Kstart from 0 to len1+len2;
int k=0;
for(j=0;j<len1;j++)
{
s[k]=s1[j];
k++;
}
for(j=0;j<len2;j++)
{
s[k]=s2[j];
k++;
}
///To place a null at the last of the concatenated string to prevent Printing Garbage value;
s[k]='\n';
cout<<s;
}
int main()
{
char s1[100];
char s2[100];
cin.getline(s1,100);
cin.getline(s2,100);
Concatenate(s1,s2);
return 0;
}
Hope it helps.
希望能帮助到你。
回答by DerfOh
for (int i = 0; i< arraySize * 2; i++)
{
if (i < aSize)
{
*(array3 + i) = *(array1 + i);
}
else if (i >= arraySize)
{
*(array3 + i) = *(array2 + (i - arraySize));
}
}
This might help you along, it doesn't require vectors. I had a similar problem in my programming class. I hope this helps, it was required that I used pointer arithmetic. This assumes that array1 and array2 are initialized dynamically to the size "aSize"
这可能会对您有所帮助，它不需要向量。我在编程课上遇到了类似的问题。我希望这会有所帮助，我需要使用指针算法。这假设 array1 和 array2 被动态初始化为大小“aSize”
回答by moldovean
Given int * arr1 and int * arr2, this program Concatenates in int * arr3 the elements of the both arrays. Unfortunately, in C++ you need to know the sizes of each arrays you want to copy. But this is no impediment to choose how many elements you want to copy from arr1 and how many from arr2.
给定 int * arr1 和 int * arr2，该程序在 int * arr3 中连接两个数组的元素。不幸的是，在 C++ 中，您需要知道要复制的每个数组的大小。但这并不妨碍您选择要从 arr1 复制多少元素以及从 arr2 复制多少元素。
#include <iostream>
using namespace std;
int main(){
int temp[] = {1,2,3,4};
int temp2[] = {33,55,22};
int * arr1, * arr2, *arr3;
int size1(4), size2(3); //size1 and size2 is how many elements you
//want to copy from the first and second array. In our case all.
//arr1 = new int[size1]; // optional
//arr2 = new int[size2];
arr1=temp;
arr2=temp2;
arr3 = new int;
//or if you know the size: arr3 = new int[size1+size2];
for(int i=0; i<size1+size2; i++){
if (i<size1)
arr3[i]=arr1[i];
else
arr3[i] = arr2[i-size1];
}
cout<<endl;
for (int i=0; i<size1+size2; i++) {
cout<<arr3[i]<<", ";
}
}
