Converting 'integer strings' to integer ar

2019-02-25 20:43发布

I'm trying to pass in an array of integers into my program. Is there a better way to convert it to integers? I'm currently getting an error: "Variable sized object may not be initialized"

for(i = 0; i < argc; i++)
{
    int arr[i] = atoi(argv[i]);
}

标签: c arrays atoi
2条回答
Luminary・发光体
2楼-- · 2019-02-25 20:51

Assuming argc and argv are the arguments passed to main, it is unlikely that argv[0] is something that you want to convert into an integer. argv[0] usually contains the name of the program.

Your code snippet is declaring an array local to the loop body. What you likely want is an array defined outside the loop body, and you want to assign to individual array elements within the loop body.

int arr[argc];
for(i = 1; i < argc; i++)
{
    arr[i] = atoi(argv[i]);
}
查看更多
对你真心纯属浪费
3楼-- · 2019-02-25 20:57

You are declaring your array arr every time you loop.

change your loop like this:

#include<stdio.h>
#include<stdlib.h>

int main(int argc, char *argv[])
{

    int arr[argc];
    int i=0;


    for(i = 0; i < argc-1; i++)
    {
       arr[i] = atoi(argv[i+1]);
       printf("arr[%d] = %d\n",i,arr[i]);
    }

    return 0;
}

Here is the output:

Sukhvir@Sukhvir-PC ~
$ gcc -Werror -Wall -g -o test test.c

Sukhvir@Sukhvir-PC ~
$ ./test 3 4 5
arr[0] = 3
arr[1] = 4
arr[2] = 5
查看更多
登录 后发表回答