How to modify the characters in a char array - arrays

Question

My intention was to use the command line to read input and store it into an array and modify the characters of the array. If the character is '1', then turn it into '0', vice versa. I successfully store the input into an array, yet failed to modify the characters of the array. If I put 0000000000000000000000000000000(32bits) into my program, the output doesn't change.
#include <stdlib.h>
#include <string.h>
int main(int argc, char *argv[]) {
char *a = argv[argc-1];
char arr[33];
size_t length = strlen(a);
for(size_t i=0;i<length;i++) {
arr[i]=a[i];
}
for(int j=0; j<32;j++) {
if(arr[j]=='0') {
arr[j]='1';
}
if(arr[j]=='1') {
arr[j]='0';
}
}
for(int k=0;k<32;k++) {
printf("%c",arr[k]);
}
}

Answer 1

If I put 0000000000000000000000000000000(32bits) into my program, the output doesn't change.
First of all, your program is not processing bits of input but it is processing the characters of input you are passing as command line argument to your program. If you want to process the bits of input, to start with, read about bitwise operators first.
Look at this for loop:
for(int j=0; j<32;j++) {
if(arr[j]=='0') {
arr[j]='1';
}
if(arr[j]=='1') {
arr[j]='0';
}
}
If current processing arr element value is '0', the second if condition in the for loop body will nullify the effect of first if condition on that arr element i.e. the arr element value will be unchanged and if the current processing arr element value is '1' than it will be set to '0'. So, after the loop, all the elements of array arr, which have original value as either '1' or '0', will set to '0' and that's the flaw in your logic.
Few suggestions:
If program is supposed to receive command line argument('s), make sure to put check on argc value. In your case, user is supposed to pass the string comprised of '0's and '1's, so you should check on argc value, something like:
if (argc != 2) {
printf ("Usage: <exe_name> <string_of_0s_and_1s>\n");
return 1;
}
Since, you do not have check on argc value in your program, if user do not pass any string from command line, the value of argc may be 1 and ,in this case, argv[argc-1] (or argv[0]) represents the program name or if the program name is not available from the host environment than argv[0][0] will be the null character. If argv[0] represents program name whose length is greater than 32 characters then your program will land up in undefined behaviour territory because there is no check on whether the command line input string length is <= 32 characters while copying it to arr buffer.
Since you are copying the input to char array (not as string) declare arr as array of 32 characters and not 33 characters. Looks like, you have assumed the input will be of size <= 32 characters. There is no problem with this assumption as long as program is running in a controlled environment where you are taking care of it. But still, by chance, if input string length is > 32 characters then your program behaviour will be undefined as it will end up processing array arr beyond it's size. You should add a check on length of input string as well:
if (length > 32) {
printf ("Input string length is >= 33 character.\n" \
"The permitted input string length is <= 32 characters.\nExiting..\n");
return 1;
}
If you want the input string to be of exact 32 characters, you can add check if (length != 32).
While copying the string, you can also add check on the characters of input string and if it contain character other than '1' and '0' then throw error message and exit.
Putting these altogether :
#include <stdio.h>
#include <string.h>
int main (int argc, char *argv[]) {
char *a = NULL;
char arr[32];
if (argc != 2) {
printf ("Usage: <exe_name> <string_of_0s_and_1s>\n");
return 1;
}
a = argv[argc-1];
size_t length = strlen(a);
if (length > 32) {
printf ("Input string length is > 32 characters.\n" \
"The permitted input string length is <= 32 characters.\nExiting..\n");
return 1;
}
for (size_t i = 0; i < length; i++) {
if ((a[i] != '1') && (a[i] != '0')) {
printf ("Found a character other than characters '1' and '0' in the input string.\n" \
"The input string should comprised of characters `1' and '0' only.\nExiting..");
return 1;
}
arr[i] = a[i];
}
for (size_t j = 0; j < length; j++) {
if (arr[j] == '0') {
arr[j] = '1';
} else {
arr[j] = '0';
}
}
for (size_t k = 0; k < length; k++) {
printf ("%c", arr[k]);
}
return 0;
}

Answer 2

you forget to give an else if condition in this part of the code.
for(int j=0; j<32;j++) {
if(arr[j]=='0') {
arr[j]='1';
}
else if(arr[j]=='1') { //This part should be else if
arr[j]='0';
}
}
In your code for the value of 0 it changed to 1. But for another if condition the value changed to 0 again.

Answer 3

To copy string you need to copy null character as well.
size_t length = strlen(a);
14 for(size_t i=0;i<=length;i++){
15 arr[i]=a[i];
But you do not need to traverse the string twice
size_t i=0;
while((arr[i]=a[i])) i++;
Your if is wrong. It should be
if(arr[j]=='0') arr[j]='1';
else arr[j]='0';