Pass string to main and break up into array - c

Question

I am passing arguments to main with this code:
#include <stdio.h>
int main(int argc, char ** argv)
{
int i = 1;
for(i = 1; i < argc; i++)
printf("%c", argv[i]);
return 0;
}
So I use ./test 218 abc 392990xFF[w2 dlx which works fine. However, the array is:
arr[1] = "218"
arr[2] = "abc"
arr[3] = "392990xFF[w2"
arr[4] = "dlx"
I want the array to be like this:
arr[0] = '2'
arr[1] = '1'
arr[2] = '8'
arr[3] = 'a'
etc...
How can I achieve this without putting a space after each digit or character?

Answer 1

The arguments passed by the run time environment to the program can be captured by main using int argc, char** argv only. If you have a need to combine them into one large array, you'll need to write the code for that, or print them one character at a time.
int main(int argc, char ** argv)
{
int i = 1;
int j;
int len;
for(i = 1; i < argc; i++)
{
len = strlen(argv[i]);
for ( j = 0; j < len; ++j )
{
printf("%c", argv[i][j]);
}
}
return 0;
}

Answer 2

First of all this is not what it will print -
arr[0] = "218"
arr[1] = "abc"
arr[2] = "392990xFF[w2"
arr[3] = "dlx"
argv[0] will store ./test. And "218" will be on index 1 thus others similarly .
And also printf("%c", argv[i]); .%c expects a char and you pass a string which is incorrect.
Solution could be -
#include <stdio.h>
int main(int argc, char ** argv)
{
int i = 1,j;
for(i = 1; i <argc; i++)
for(j=0;argv[i][j]!='\0';j++)
printf("%c\n", argv[i][j]);
return 0;
}

Answer 3

Rather than a for loop, you can also simply use pointers and while loops instead. There are generally many ways to solve problems in C:
#include <stdio.h>
int main (int argc, char **argv) {
int i = 1;
int j = 0;
while (i < argc) {
char *p = argv[i];
while (*p) {
printf (" arr[%2d] = \"%c\"\n", j++, *p);
p++;
}
i++;
}
return 0;
}
Output
$ ./bin/argvchars 218 abc 392990xFF[w2 dlx
arr[ 0] = "2"
arr[ 1] = "1"
arr[ 2] = "8"
arr[ 3] = "a"
arr[ 4] = "b"
arr[ 5] = "c"
arr[ 6] = "3"
arr[ 7] = "9"
arr[ 8] = "2"
arr[ 9] = "9"
arr[10] = "9"
arr[11] = "0"
arr[12] = "x"
arr[13] = "F"
arr[14] = "F"
arr[15] = "["
arr[16] = "w"
arr[17] = "2"
arr[18] = "d"
arr[19] = "l"
arr[20] = "x"

Answer 4

Determine the total number of characters in all of the strings, then allocate a new character array of that length, and then copy the input characters into the new array.
The last part could take advantage of the sizes you gather in the first part: have an outer loop over all the argument strings, with an inner loop over the characters in each string.
EDIT: Now that I'm not on a mobile device, here's the above in code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main(int argc, char ** argv)
{
//For storing sizes of each input string
int *arg_chars;
//Where the individual characters are stored
char *stored_chars;
/* Determine total number of characters, and store
characters in each word for later re-use */
arg_chars = malloc(argc * sizeof(int));
int total_chars = 0;
//Loop starts at 1 since we don't care about arg 0
for(int i=1; i<argc; i+=1)
{
arg_chars[i] = strlen(argv[i]);
total_chars += arg_chars[i];
printf("Word %d is %d long\n", i, arg_chars[i]);
}
/* Load argument characters into the stored_chars array */
stored_chars = malloc(total_chars * sizeof(char));
int current_char = 0;
//Loop starts at 1 to exclude the program name (arg 0)
for(int i = 1; i < argc; i+=1)
{
printf("Scanning word %d (%s):\n", i, argv[i]);
for(int j = 0; j < arg_chars[i]; j+=1)
{
stored_chars[current_char] = argv[i][j];
printf(" Stored letter %d `%c` (letter %d of word %d)\n", current_char, argv[i][j], j, i);
current_char += 1;
}
}
/* Demonstrate that it's all loaded and accessible in any order */
for(int i=total_chars-1; i >= 0; i-=1)
{
printf("stored_chars[%d] = `%c`\n", i, stored_chars[i]);
}
return 0;
}
Output:
Word 1 is 3 long
Word 2 is 3 long
Word 3 is 12 long
Word 4 is 3 long
Scanning word 1 (218):
Stored letter 0 `2` (letter 0 of word 1)
Stored letter 1 `1` (letter 1 of word 1)
Stored letter 2 `8` (letter 2 of word 1)
Scanning word 2 (abc):
Stored letter 3 `a` (letter 0 of word 2)
Stored letter 4 `b` (letter 1 of word 2)
Stored letter 5 `c` (letter 2 of word 2)
Scanning word 3 (392990xFF[w2):
Stored letter 6 `3` (letter 0 of word 3)
Stored letter 7 `9` (letter 1 of word 3)
Stored letter 8 `2` (letter 2 of word 3)
Stored letter 9 `9` (letter 3 of word 3)
Stored letter 10 `9` (letter 4 of word 3)
Stored letter 11 `0` (letter 5 of word 3)
Stored letter 12 `x` (letter 6 of word 3)
Stored letter 13 `F` (letter 7 of word 3)
Stored letter 14 `F` (letter 8 of word 3)
Stored letter 15 `[` (letter 9 of word 3)
Stored letter 16 `w` (letter 10 of word 3)
Stored letter 17 `2` (letter 11 of word 3)
Scanning word 4 (d1x):
Stored letter 18 `d` (letter 0 of word 4)
Stored letter 19 `1` (letter 1 of word 4)
Stored letter 20 `x` (letter 2 of word 4)
stored_chars[20] = `x`
stored_chars[19] = `1`
stored_chars[18] = `d`
stored_chars[17] = `2`
stored_chars[16] = `w`
stored_chars[15] = `[`
stored_chars[14] = `F`
stored_chars[13] = `F`
stored_chars[12] = `x`
stored_chars[11] = `0`
stored_chars[10] = `9`
stored_chars[9] = `9`
stored_chars[8] = `2`
stored_chars[7] = `9`
stored_chars[6] = `3`
stored_chars[5] = `c`
stored_chars[4] = `b`
stored_chars[3] = `a`
stored_chars[2] = `8`
stored_chars[1] = `1`
stored_chars[0] = `2`