C语言 两个数组的总和
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Sum of two arrays
提问by captain
The exercise says "Make a function with parameters two int arrays and k which is their size. The function should return another array (size k) where every element of it is the sum of the two arrays of the same position. That's what I wrote, but it crashes. Do I have to do it with pointers?
练习说“创建一个函数，参数为两个 int 数组，k 是它们的大小。该函数应该返回另一个数组（大小为 k），其中每个元素都是相同位置的两个数组的总和。这就是我写的，但它崩溃了。我必须用指针来做吗？
#include <stdio.h>
#include <stdlib.h>
void sumarray(int k,int A[k],int B[k]){
int sum[k],i;
for(i=0;i<k;i++){
sum[i]=A[i]+B[i];
printf("sum[%d]=%d\n",i,sum[i]);}
}
main(){
int i,g,a[g],b[g];
printf("Give size of both arrays: ");
scanf("%d",&g);
for(i=0;i<g;i++){
a[i]=rand();
b[i]=rand();
}
sumarray(g,a,b);
system("pause");
}
Example: If i have A={1,2,3,4} and B={4,3,2,1} the program will return C={5,5,5,5).
示例：如果我有 A={1,2,3,4} 和 B={4,3,2,1}，程序将返回 C={5,5,5,5)。
回答by Oliver Charlesworth
This:
这个：
int i,g,a[g],b[g];
causes undefinedbehaviour. The value of gis undefined upon initialisation, so therefore the length of aand bwill be undefined.
导致未定义的行为。的值g在初始化未定义，所以因此的长度a和b将是不确定的。
You probably want something like:
你可能想要这样的东西：
int i, g;
int *a;
int *b; // Note: recommend declaring on separate lines, to avoid issues
scanf("%d", &g);
a = malloc(sizeof(*a) * g);
b = malloc(sizeof(*b) * g);
...
free(a);
free(b);
回答by The GiG
Its impossible to first do a[g]when dynamically getting g.
它不可能先做a[g]时动态获取g。
Your first lines in main should be:
main 中的第一行应该是：
int i,g;
int *a,*b;
printf("Give size of both arrays: ");
scanf("%d",&g);
a = (int *)malloc(g*sizeof(int));
b = (int *)malloc(g*sizeof(int));
回答by Priyank
change the function summary signature (the definition part of the declaration) to this and try it out:
将函数摘要签名（声明的定义部分）改成这样，试试看：
void sumarray(int k,int* A,int* B){
void sumarray(int k,int* A,int* B){
回答by Mahesh
int sum[k] ;
kis a variable but the size of the array should be a constant.
k是一个变量，但数组的大小应该是一个常量。
The function should return another array (size k) ...
该函数应返回另一个数组（大小为 k）...
But the function you wrote returns voidwhich is clearly wrong.
但是您编写的函数返回void显然是错误的。
Do I have to do it with pointers?
我必须用指针来做吗？
Yes.
是的。
回答by Eric Conner
One issue is that you've attempted to declare dynamically sized arrays on the stack (e.g. a[g]). You need to declare pointers for each array and then dynamically allocate your variable sized array once you've read in the value of g.
一个问题是您试图在堆栈上声明动态大小的数组（例如a[g]）。您需要为每个数组声明指针，然后在读取g.
