Question

int main(void) {
    int no, flimit, i, pos_of_fval, freq_of_fval, n, j;
    long long int max;
    scanf("%d %d", &n, &flimit);
    char list[n][21];
    long long int fval[n];
    char disp[flimit][21];
    for (i = 0; i < n; i++)
        scanf("%s  %lld", list[i], &fval[i]);
    for (i = 0; i < flimit; i++) {
        max = 0;
        for (j = 0; j < n; j++)
            if (fval[j] > max) {
                max = fval[j];
                pos_of_fval = j;
            }
        freq_of_fval = freq(fval, max, n);
        if (freq_of_fval == 1) {
            strcpy(disp[i], list[pos_of_fval]);
            fval[pos_of_fval] = 0;
        } else
            lexical(list, disp, fval, i, n, freq_of_fval, max);
    }
    for (i = 0; i < flimit; i++)
       printf("%s\n", disp[i]);

    return 0;
}

int freq(long long fval[], long long max, int n) {
    int count = 0, i;
    for (i = 0; i < n; i++)
        if (fval[i] == max)
    count++;
    return count;
}

void lexical(char list[][21], char disp[][21], long long fval[],
             intn dispos ,intn, int freq_of_fval, long long max) {

    int a[freq_of_fval], i, j, apos = 0, temp;

    for (j = 0; j < freq_of_fval; j++) {
        for (i = 0; i < n; i++)
            if (fval[i] == max)
                a[apos++] = i;
    }

    for (i = 0; i < freq_of_fval - 1; i++) {
        for (j = 0; j < freq_of_fval - 1 - i; j++) {
            if (strcmp((list[a[j]]), list[a[j + 1]]) > 1) {
                temp = a[j];
                a[j] = a[j + 1];
                a[j + 1] = temp;
            }
        }
    }
    strcpy(disp[dispos], list[a[0]]);
}

错误：

warning: passing argument 1 of 'lexical' from incompatible pointer type
lexical(list,disp,fval,i,n,freq_of_fval,max);  
            note: expected 'char \*' but argument is of type 'char (\*)[21]'
     void lexical(char list[],char disp[][21],long long fval[], int dispos, int

Answer 1

您好，欢迎来到C编码的精彩世界 - 您需要在阵列长度上保持紧张的统治。

此代码存在以下问题 - 首先，您要声明大小未知的数组。例如在 -

char list[n][21];

这样的行只能在C99中工作 - 否则你不能声明一个可变大小的数组。但即使在这种情况下你也可以这样做 - 在你的情况下，在你使用它之前从未设置过n。

通常，在C中 - 您必须选择： 1.你事先知道你的数组大小，然后在分配和函数签名中使用它

例如，如果你有20x21矩阵，那么比主代码：

char list[20][21]

然后在函数

中

void lexical(char list[20][21],...)

或

void lexical(charlist[][],...)

如果您事先不知道大小，则可以使用malloc

：

//N is how many rows you have
char **list  = malloc(sizeof (char *) * N);
if (list  )
{
  for (i = 0; i < N; i++)
  {
    a[i] = malloc(sizeof(char) * 21);
  }
}

然后在函数中： void lexical（char ** list，...）

Answer 2

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

void lexical(char list[][21], char disp[][21], long long fval[],
             int dispos ,int n, int freq_of_fval, long long max);
int freq(long long fval[], long long max, int n) ;

int main(void) {

    int no, flimit, i, pos_of_fval, freq_of_fval, n, j;
    long long int max;

    printf("请输入n 和 flimit\n");
    scanf("%d %d", &n, &flimit);

    char list[n][21];
    long long int fval[n];
    char disp[flimit][21];
    for (i = 0; i < n; i++) {

        printf("字符串 和 long long数据\n");
        scanf("%s  %lld", list[i], &fval[i]);
    }

    for (i = 0; i < flimit; i++) {

        max = 0;
        for (j = 0; j < n; j++)
        {
            if (fval[j] > max) {
                max = fval[j];
                pos_of_fval = j;
            }
        }

        freq_of_fval = freq(fval, max, n);

        if (freq_of_fval == 1) {
            strcpy(disp[i], list[pos_of_fval]);
            fval[pos_of_fval] = 0;
        } else {
            lexical(list, disp, fval, i, n, freq_of_fval, max);
        }
    }

    for (i = 0; i < flimit; i++)
        printf("%s\n", disp[i]);

    return 0;
}

// 计算数组中最大值有多少个(相同的最大值)
int freq(long long fval[], long long max, int n) {

    int count = 0, i;
    for (i = 0; i < n; i++)
        if (fval[i] == max)
            count++;
    return count;
}

void lexical(char list[][21], char disp[][21], long long fval[],
             int dispos ,int n, int freq_of_fval, long long max) {

    int a[freq_of_fval], i, j, apos = 0, temp;

    for (j = 0; j < freq_of_fval; j++) {
        for (i = 0; i < n; i++)
            if (fval[i] == max)
                a[apos++] = i;
    }

    for (i = 0; i < freq_of_fval - 1; i++) {
        for (j = 0; j < freq_of_fval - 1 - i; j++) {
            if (strcmp((list[a[j]]), list[a[j + 1]]) > 1) {
                temp = a[j];
                a[j] = a[j + 1];
                a[j + 1] = temp;
            }
        }
    }
    strcpy(disp[dispos], list[a[0]]);
}

如何传递任何类型的2d数组

2 个答案: