printing a string in c characer by character - c

Question

i know , that my question is unusual behaviour for coding , but i just do it , to understand C language in deep , so i ask user to enter a string and i want to print this string character by character , before doing this , i save the string in memory , which it's address saved in 'x' , can i print this string character by character , by this way ?
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
int main(int argc ,char *argv[20])
{
char x ;
strcpy(&x,argv[1]);
printf("%s and it's length is %d : \n" , &x , strlen(&x));
int i ;
for(i=0 ; i < strlen(&x) ; i++)
{
printf(" hi : %c\n" , x[i]);
/* also i try '%x[i]' but it's give me another error
so how i do this ??*/
}
return 0;
}

Answer 1

The basic idea is correct, but there are some problems in the code.
x is only a single character. You´ll need an array of it.
Then replace every &x with x, because the array pointer itself is already an address.
And another thing: char *argv[20].
Why 20? Why not char *argv[] ?

Answer 2

So, starting from the beginning:
int main(int argc ,char *argv[20])
^^^^
This does nothing; whenever you declare a function parameter as T a[N] or T a[], it's intepreted as T *a; the traditional way to write this is as either
int main( int argc, char *argv[] )
or
int main( int argc, char **argv )
The second case more accurately reflects what's really happening; argv is a pointer to a sequence of pointers to char. The value of argv[argc] is NULL.
{
char x ;
strcpy(&x,argv[1]);
This is a problem; by declaring x as char, you've only set aside enough storage to hold a single character value1, but you're trying to use it to store an array of character values; that isn't going to work, and you're going to wind up clobbering whatever memory immediately follows x, which may or may not lead to a crash. C doesn't do any bounds checking for you; when you give strcpy an address, it will happily copy the given string to the storage following that address without making sure that storage is valid.
You have two options at this point: you can either declare x as an array of char:
char x[SOME_SIZE]; // where SOME_SIZE is large enough to handle your
// largest expected input
and use strcpy as follows:
strcpy( x, argv[1] );
or you can simply declare x as a pointer to char:
char *x;
and assign it to point to the beginning of argv[1]:
x = argv[1];
Either way, the rest of the code becomes
printf("%s and it's length is %d : \n" , x , strlen(x)); // no & before x
int i ;
for(i=0 ; i < strlen(x) ; i++)
{
printf(" hi : %c\n" , x[i]);
}
return 0;
}
Note that a much simpler way of doing this would be
int main( int argc, char **argv )
{
char *p = argv[1];
if ( p ) // argv[1] is not NULL
{
while ( *p ) // argv[1][i] is not 0
printf( "hi: %c\n", *p++ );
}
return 0;
}
1. Which is why x[i] wasn't working for you, since x is not an array or pointer expression

Answer 3

deviantfan is right that you are overwriting something in memory that does not belong to the variable. Here is one of the ways to accomplish your task without any memory-editing voodoo:
#define ARGV_SIZE 20
int main(int argc ,char *argv[ARGV_SIZE])
{
char x[ARGV_SIZE];
int argvLength;
strcpy(x, argv[1]);
argvLength = strlen(x);
printf("%s and it's length is %d : \n" , x , argvLength);
for (int i = 0 ; i < argvLength; i++)
{
printf(" hi : %c\n" , x[i]);
}
return 0;
}
or without the x variable you can do this:
int main(int argc ,char *argv[])
{
int argvLength = strlen(argv[1]);
printf("%s and it's length is %d : \n" , argv[1] , argvLength);
for (int i = 0 ; i < argvLength; i++)
{
printf(" hi : %c\n" , argv[1][i]);
}
return 0;
}

Answer 4

try this
#include<stdio.h>
#include<string.h>
int main(int argc ,char *argv[20]){
char y = 'Y';
char x ;
char z = 'Z';
strcpy(&x, "This program is correct!!!");
printf("%s and it's length is %d : \n" , &x , strlen(&x));
printf("%c%c%c", x,y,z);//print TYZ ?
return 0;
}