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c - How to turn a horizontal matrix vertical?

问题描述

I'm converting specific hexadecimals based on the entered character to decimals and then to binary. I'm trying to print a matrix of the 1s and 0s in the shape of the entered character. When I print the first character it outputs sideways. How can I turn it vertical?

I tried finding a way to print just the first digit of each binary on one row of columns, the second digit on the 2nd row of columns, and so on, but came up with nothing. The code requires the use of an external file with all of the hexadecimals. When pulled from the file, the hexadecimals for A are in decimal form: 126, 17, 17, 17, 126. I then convert them to a binary integer: 01111110, 00010001, 00010001, 00010001, and 01111110. And print them out in a 5x7 matrix. It should produce a vertical A but instead produces a horizontal A.

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include "font5x7.h"

void DecToBin (int newBin[], char userMessage[], int k) {
int i = 0;
int j = 0;
int m = 0;

for (i = 0; i < 1; ++i) {
    while (m < 8) {
    newBin[j] = Font5x7[k] % 2;
    Font5x7[k] = Font5x7[k] / 2;
    ++j;
    ++m;
    }
}
}

int main (void) {

const int INPUT_STR_SIZE = 40;
char userMessage[INPUT_STR_SIZE];
int i = 0;
int j = 0;
int k = 0;

const int BIN_MAX = 8;
int binNum[BIN_MAX];

printf("Enter message:\n");
fgets(userMessage, INPUT_STR_SIZE, stdin);



//Iterate through DecToBin function for bin equivalents
for (i = 0; i < strlen(userMessage); ++i) {
    if (userMessage[i] == 'A') {
        for (k = 165; k < 170; ++k) {
            DecToBin(binNum, userMessage, k);
                for (j = 7; j >= 0; --j) {
                    printf("%d", binNum[j]);
                }
                    printf("\n");
        }
    }
}






return 0;

}

标签： carraysloopsmatrixorientation

解决方案

If I well understand the horizontal A with pixels from 126, 17, 17, 17, 126 :

01111110
00010001
00010001
00010001
01111110

must produce a vertical A turning left (missing 0 not present here):

01110
10001
10001
10001
11111
10001
10001
00000

That means : (the upper digit is the index in the array, the lower the bit rank)

07 06 05 04 03 02 01 00
17 16 15 14 13 12 11 10
27 26 25 24 23 22 21 20
...
n7 n6 n5 n4 n3 n2 n1 n0

have to produce

00 10 20 30 .. n0
01 11 21 31 .. n1
02 12 22 32 .. n2
...
07 17 27 37 .. n7

A proposal to do that :

#include <stdio.h>
#include <string.h>

void turnLeft(const unsigned char * v, unsigned char r[8], size_t sz)
{
  memset(r, 0, sizeof(r));

  size_t j = 0;

  for (unsigned m = 1; m != (1 << 8); m <<= 1) {
    for (size_t i = 0; (i != sz) && (i != 8); ++i) {
      if (v[i] & m)
        r[j] |= (1 << (7 - i));
    }
    j += 1;
  }
}

void pr(const unsigned char * v, size_t sz)
{
  for (size_t i = 0; i != sz; ++i) {
    unsigned char c = v[i];

    for (unsigned m = (1 << 7); m != 0; m >>= 1)
      putchar('0' + ((c & m) != 0));

    putchar('\n');
  }
}

int main()
{
  const unsigned char v[] = { 126, 17, 17, 17, 126 };
  unsigned char r[8];

  pr(v, sizeof(v));
  putchar('\n');

  turnLeft(v, r, sizeof(v));
  pr(r, 8);
}

Compilation and execution :

vxl15036 /tmp % gcc -std=c99 -pedantic -Wextra a.c
vxl15036 /tmp % ./a.out
01111110
00010001
00010001
00010001
01111110

01110000
10001000
10001000
10001000
11111000
10001000
10001000
00000000

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