How to use 2D array of characters name as a pointer in C

5 Answers

0 votes
#include <stdio.h>

int main(int argc, char **argv)
{ 
	char arr[2][3] = {{'a','b','c'},{'d','e','f'}}; 
	
    printf("%s\n", *arr); 

	return 0; 
}   
 

/*
run:
 
abcdef
 
*/

 



answered Jan 19, 2019 by avibootz
0 votes
#include <stdio.h>

int main(int argc, char **argv)
{ 
	char arr[2][3] = {{'a','b','c'},{'d','e','f'}}; 
	
    printf("%s\n", *(arr + 1)); 

	return 0; 
}   
 

/*
run:
 
def
 
*/

 



answered Jan 19, 2019 by avibootz
0 votes
#include <stdio.h>

int main(int argc, char **argv)
{ 
	char arr[2][3] = {{'a','b','c'},{'d','e','f'}}; 
	
    printf("%c\n", *(*(arr + 0))); 
    printf("%c\n", *(*(arr + 1))); 
	
	return 0; 
}   
 

/*
run:
 
a
d
 
*/

 



answered Jan 19, 2019 by avibootz
0 votes
#include <stdio.h>

int main(int argc, char **argv)
{ 
	char arr[2][3] = {{'a','b','c'},{'d','e','f'}}; 
	
    printf("%c\n", *(arr + 0)[0]); 
    printf("%c\n", *(arr + 1)[0]); 
	
	return 0; 
}   
 

/*
run:
 
a
d
 
*/

 



answered Jan 19, 2019 by avibootz
0 votes
#include <stdio.h>

int main(int argc, char **argv)
{ 
	char arr[2][3] = {{'a','b','c'},{'d','e','f'}}; 
	
    printf("%c\n", **(arr + 0) + 1); 
    printf("%c\n", **(arr + 1) + 1); 
	
	return 0; 
}   
 

/*
run:
 
b
e
 
*/

 



answered Jan 19, 2019 by avibootz

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