C language: How do I convert a string parameter from a command line into an array of individual characters?

Question

here is a part of my code:

int main(int argc, char*argv[])  
{
 //This section shows the statistic of the program
int a;

printf("The number of arguments is :%d \n", argc);

for(a=0; a<argc; a++)
{
        printf("argc %d is %s \n",a, argv[a]);
}

Based on the code above, the second command line argument argv[1] will be a .txt file.

The third command line argument argv[2] will be entered as a string such as "monkey".

I need to know how to store the second command line argument "monkey" as an array of individual character so it will look somethings like this:

char key[6];
     key[0]= 'm';
     key[1]= 'o';
     key[2]= 'n';
     key[3]= 'k';
     key[4]= 'e';
     key[5]= 'y';

@Mr.goosberry from what i see, argv[2] = 'monkey'. So what do I do if i want to print each character in that word individually? — Uncle Zhang He or ZhangHim Her, Oct 01 '16 at 04:38
"from what i see, ... So what do I do ..." - Simply see a good C book before asking such questions. @Mr.goosberry is completely right. — too honest for this site, Oct 01 '16 at 04:48
"the second command line argument argv[1] will be a .txt file." - Unless you `echo` the whole contents as argument, you likely just pass **the name** of a file. — too honest for this site, Oct 01 '16 at 04:51
@UncleZhangHeorZhangHimHer This is where you could start - http://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list. — Christian Dean, Oct 01 '16 at 04:54

msc · Accepted Answer · 2016-10-01T05:16:35.087

-1

Try this:

#include <stdio.h>
#include <string.h>


int main(int argc, char*argv[])
{
        //This section shows the statistic of the program
        int a;

        printf("The number of arguments is :%d \n", argc);

        for(a=0; a<argc; a++)
        {
                printf("argc %d is %s \n",a, argv[a]);
        }

        size_t len = strlen(argv[2]);

        int key[10];
        int i = 0;
        for(i = 0; i < len; i++)
        {
                key[i] = argv[2][i];
        }

        for (i = 0; i < len; i++)
        {
                printf("key[%d] = %c\n", i, key[i]);
        }
}

edited Oct 01 '16 at 05:16

answered Oct 01 '16 at 04:40

msc

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No help. Code-only answer – too honest for this site Oct 01 '16 at 04:49
Among other errors, If argv[2] takes >10 bytes, this will write past the end of the array, causing a serious memory error. It's never a good idea to have a fixed-size array for user input. – AlexPogue Oct 01 '16 at 04:51
2

Thank you M.S Chaudhari for actually trying to help. – Uncle Zhang He or ZhangHim Her Oct 01 '16 at 04:55
This `= strlen(argv[2]+1);` should be `= strlen(argv[2]);` along with those `for (i = 0; i <= len; i++)` becoming `for (i = 0; i < len; i++)`. Also `strlen()` returns `size_t` not `int`. – alk Oct 01 '16 at 05:12

C language: How do I convert a string parameter from a command line into an array of individual characters?

1 Answers1