c - Multidimensional Array Printing Gives Something Different -

#include <stdio.h>  int main(int argc, char **argv) {  int x[5][5] = {{0,1,2,3,4},{5,6,7,8,9}};  (int = 0; < 4; i++) {     (int j = 0; < 4; i++) {         printf("%d\n", x[i][j]); } }  return 0; } 

this supposed give me: 0 1 2 3 4 5 6 7 8 9

and result is: 0 5 0 0

and then, when change them chars:

#include <stdio.h>  int main(int argc, char **argv) {  char x[5][5] = {{'0','1','2','3','4'},{'5','6','7','8','9'}};  (int = 0; < 4; i++) {     (int j = 0; < 4; i++) {         printf("%d\n", x[i][j]); } }  return 0; } 

i 48 53 0 0.

why? code pretty clear me, seems happening obscure on background (or brain works in "pythonic" way...)

like @keith thompson said, didn't initialize 25 elements. remaining elements initialized 0. result, data stored way:

  0 | 1 | 2 | 3 | 4     // row 0   5 | 6 | 7 | 8 | 9     // row 1   0 | 0 | 0 | 0 | 0     // row 2   0 | 0 | 0 | 0 | 0     // row 3   0 | 0 | 0 | 0 | 0     // row 4 

you initialized first 2 rows of array. reason why got 0 5 0 0 output.

to correct output, need loop through first 2 rows of array.

steps fix :

first

you need change first for-loop from:

for (int = 0; < 4; i++) 

to

for (int = 0; < 2; i++) 

reason: because want loop through first 2 rows (row 0 , row 1).

second

you need change second for-loop from:

for (int j = 0; < 4; i++) 

to:

for (int j = 0; j < 5; j++) 

reason: need variable j loop condition, not i. need change 4 5; because in array, index starts 0. so, indexes of each columns 0,1,2,3,4 , respectively.

to sum up, code supposed be...

for (int = 0; < 2; i++)  {     (int j = 0; j < 5; j++)      {         printf("%d\n", x[i][j]);      }  } 

additionally, when changing char x[5][5], need use %c, print character, instead of %d.

moreover, reason getting weird numbers 48 because when print char %d, printing ascii values. referring ascii table below, number 48 represents character '0'.

image http://www.asciitable.com/