How can I code a program that will encode a text written only in capital letters according to the text below?

Question

 1 2 3 4 5 6 1 A B C D E F 2 G H I J K L 3 M N O P R S 4 T U V Y Z W 5 X 1 2 3 4 5 6 6 7 8 9 ? ! 7 - + *

I want to do that.

HELLO 1234 equivalent -> 221526263352535455

It is output that I got.

636465

I started with a simple logic, but could not continue.

My codes.

#include <stdio.h> #include <string.h> int main(){ int i,j; char string[20]; char b[7][6]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','R','S','T','U' ,'V','Y','Z','W','X','1','2','3','4','5','6','7','8','9','?','!','-','+','*'}; printf("please enter an input"); scanf("%s",&string); for(i=0;i<7;i++){ for(j=0;j<6;j++){ if(string[i]==b[i][j]) printf("%d%d",i,j); } } return 0; }

I've made a correction but still can't get the output I want. — ErenB
– ErenB, Commented Dec 24, 2020 at 16:54
it would be easier if you change the b[][] to be indexed by letters, like b['H'] = 22 ... — Hamza Abdaoui
– Hamza Abdaoui, Commented Dec 24, 2020 at 16:54
Yes, but how else can I print the numbers in the table side by side? For example, how can I print the row value of matrix 1 and column value 5 for the letter 'E'? — ErenB
– ErenB, Commented Dec 24, 2020 at 16:59

Krishna Kanth Yenumula · Accepted Answer · 2020-12-24 17:36:32Z

I created an char array of 256 elements, to account for all ASCII characters. I stored corresponding value of the character in the array. For example: A-->11, alpha_num['A'] = 11. Similarily, I assigned all the remaining characters.

#include <stdio.h> #include <string.h> int main() { int i = 1, j = 1; int k = 0; char alpha_num[256]; char string[20]; printf("Please enter an input\n"); scanf("%s", string); for (char c = 'A'; c <= 'Z'; c++) { if (j <= 6 && i <= 7) { alpha_num[c] = i * 10 + j; j++; } if (j > 6) { j = 1; i++; } if (i > 7) i = 1; } i = 5; j = 2; for (char c = '1'; c <= '9'; c++) { if (j <= 6 && i <= 7) { alpha_num[c] = i * 10 + j; j++; } if (j > 6) { j = 1; i++; } if (i > 7) i = 1; } alpha_num['?'] = 65; alpha_num['!'] = 66; alpha_num['-'] = 71; alpha_num['+'] = 72; alpha_num['*'] = 73; for (int i = 0; i < strlen(string); i++) { printf("%d", alpha_num[string[i]]); } return 0; }

The output is :

Please enter an input HELLO1234 221526263352535455

abelenky · Accepted Answer · 2020-12-24 19:23:50Z

I went for short and simple:

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#include <stdio.h> #include <string.h> void encode(char* input, char* output, char* matrix) { while(*input) { size_t idx = strchr(matrix, *input++) - matrix; *output++ = '1' + idx / 6; *output++ = '1' + idx % 6; } *output = 0; } int main(void) { char matrix[]="ABCDEFGHIJKLMNOPRSTUVYZWX123456789?!-+* "; char input[50]; printf("please enter an input"); fgets(input, 50, stdin); *strchr(input, '\n') = '\0'; size_t len = strlen(input); char output[2*len +1]; encode(input, output, matrix); printf("\nEncoded String: %s\n", output); return 0; }

Jonathan Leffler · Accepted Answer · 2020-12-24 17:20:53Z

You can iterate over string characters and check if the current char is found in b. If it's found you can just print the row+1 and the column+1:

#include <stdio.h> #include <string.h> char b[7][6]={'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','R','S','T','U' ,'V','Y','Z','W','X','1','2','3','4','5','6','7','8','9','?','!','-','+','*'}; void findRowCol(char c,int *row,int *col){// change row,col to the founded index, if not found to -1 *row=-1; *col=-1; for(int i=0;i<7;i++){ for(int j=0;j<6;j++){ if(c==b[i][j]){ *row=i; *col=j; } } } } int main() { char string[20] = "HELLO1234"; for(int i=0;i<strlen(string);i++){ int row,col; findRowCol(string[i],&row,&col); if(row!=-1) printf("%d%d",row+1,col+1); } }

Eden Moshe · Accepted Answer · 2020-12-24 17:21:12Z

 #include <stdio.h> struct CodeList { int code; int key; }; int main() { struct CodeList list[6*7]; int i = 1; int y =1; int x =1; for (char c = 'A'; c <='Z'; c++){ if (x > 6) { x =1; y++; } if (y > 7) { y =1; } list[i].key = c; list[i].code = (y * 10) + x; i++; x++; } //search char s = 'D'; for (i =0; i< 6*7; i++){ if(s == list[i].key){ printf("code %i", list[i].code);// 14 } } }

By the way, you ignored the letter Q.

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