c - 如何测试文本文件中的十六进制字符串？

Question

bool testMagicSquare(string str) {
  int length, sum = 0;
  length = strlen(str);

  for (int i = 0; i < length; i++) {
    sum = sum + str[i];
  }
  if (sum == 120) {
    return 1;
  }

  else return 0;
}

我的任务是从文本文件中获取一个十六进制平方，测试它是否是一个“幻方”。如果是，请在另一个文本文件中打印该正方形以保存它。获取文本文件并打印到文件正在工作，但我坚持测试。

有什么想法吗？

十六进制字符串示例:8 3 4 f 5 e 9 2 b 0 7 c 6 d a 1

Answer 1

试试这个代码。它仅适用于一位十六进制数字!

#include <stdio.h>
#include <ctype.h>

#define DIMS 4
#define COLS DIMS
#define ROWS DIMS

int convertChar(unsigned char v);
int checkAndComputeSums(int *s , unsigned char *ms, int dim);

enum STYPE {
    SUMROW,
    SUMCOL,
    SUMDIAG,

    //-------------------

    SUMCNT
};

unsigned char msqr[COLS][ROWS] = {
    {'8','3','4','F'},
    {'5','E','9','2'},
    {'B','0','7','C'},
    {'6','D','A','1'}
};

int sum[DIMS][SUMCNT];

const char * label[SUMCNT] = {
    "ROWS","COLS","DIAG"
};

int convertChar(unsigned char v)
{
    int x;

    x = toupper((int)v);
    x -= 48; /* 48 is the ASCII code of the char 0 */
    if (x>9) x-=7;

    return x;
}

int checkAndComputeSums(int *s , unsigned char *ms, int dim)
{
    int i,j,ok=1;

    /* The sum are cleared */
    for(i=0;i<dim;i++) {
        for(j=0;j<SUMCNT;j++) {
            s[i*SUMCNT+j] =0;
        }
    }

    for(i=0;i<dim;i++) {
        for(j=0;j<dim;j++) {
            s[i*SUMCNT+SUMROW]+=convertChar(ms[i*dim+j]);
            s[i*SUMCNT+SUMCOL]+=convertChar(ms[j*dim+i]);

            if (i==j)
                s[SUMDIAG]+=convertChar(ms[i*dim+j]);

            if (i==dim - j - 1)
                s[SUMDIAG+SUMCNT]+=convertChar(ms[i*dim+j]);

        }

        if (s[i*SUMCNT+SUMROW]!=s[SUMROW] || s[i*SUMCNT+SUMCOL]!=s[SUMROW])
            ok=0;
    }

    if (s[SUMDIAG]!=s[SUMROW] || s[SUMDIAG+SUMCNT]!=s[SUMROW])
        ok=0;

    return ok;
}

int main(void)
{
    int i,j;

    i=checkAndComputeSums(sum[0],msqr[0],DIMS);
    printf("The check was %s\n",(!i)?"KO":"OK");

    for (j=SUMROW;j<SUMCNT;j++) {
        printf("%s: ",label[j]);
        for(i=0;i<DIMS-((j==SUMDIAG)?2:0);i++) {
            printf("%d ",sum[i][j]);
        }
        puts("");
    }

    return 0;
}