Question

我正在扫描输入文件中的字符，并尝试打印它们在输出文件中出现的次数。

我已经尝试为循环专门创建一个新的文件流，因为它已在上一部分代码中使用。我还尝试过将long long int强制转换为数组和print语句，这只会给我更大的数字。

  FILE* input;
  FILE* output;
  input  = fopen("input.txt", "r");
  output = fopen("report.txt", "w");
  ...

  char letterTemp;
  float letterMax;
  int upperCounter = 0;
  int lowerCounter = 0;
  char alphaChars[26];
  int charCounter[26];

  for (char c = 'a'; c <= 'z'; ++c) {
    alphaChars[c - 'a'] = c; 
  }

  for (int i = 0; i < 26; i++) {
    fopen("input.txt", "r");

    while ((letterTemp = fgetc(input)) != EOF) {
      if (alphaChars[i] == letterTemp) {
        lowerCounter++;
        charCounter[i]++;
      }
      else if (toupper(alphaChars[i]) == letterTemp) {
        upperCounter++;
        charCounter[i]++;
      }
    }
    fclose(input);
  }
  letterMax = lowerCounter + upperCounter;

  fprintf(output, "LETTER STATISTICS\n\nCategory                How many in file    %% of all letters\n----------------------------------------------------------------------\nUppercase%31d%20.2f %%\nLowercase%31d%20.2f %%\n", upperCounter, 100 * upperCounter / letterMax, lowerCounter, 100 * lowerCounter / letterMax);

  for (int i = 0; i < 25; i++) {
    fprintf(output, "%c%39d%20.2f %%\n", alphaChars[i], charCounter[i], 100 * charCounter[i] / letterMax);
  }

哪个会给我...

LETTER STATISTICS

Category                How many in file    % of all letters
----------------------------------------------------------------------
Uppercase                              9               12.16 %
Lowercase                             65               87.84 %
a                             2080375142         25393028.00 %
b                                  32766            44278.38 %
c                             2080375153         25393042.00 %
d                                  32767            44279.73 %
e                             -677299551         13371914.00 %
f                                  32734            44235.14 %
g                                      0                0.00 %
h                                      2                2.70 %
i                                      7                9.46 %
j                                      0                0.00 %
k                                      0                0.00 %
l                                      7                9.46 %
m                                      2                2.70 %
n                                      1                1.35 %
o                                      7                9.46 %
p                                      1                1.35 %
q                             2081514112         26932176.00 %
r                                  32770            44283.79 %
s                             -677300459         13370687.00 %
t                                  32735            44236.49 %
u                                      2                2.70 %
v                                      1                1.35 %
w                                      2                2.70 %
x                                      1                1.35 %
y                                      5                6.76 %

这是怎么回事？

Answer 1

由于原始帖子使用toupper，因此请使用tolower islower和isupper以避免多次读取文件并循环匹配字符。

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

int main ( ) {
    FILE* input = NULL;
    FILE* output = NULL;
    int letterTemp;
    float letterMax;
    int upperCounter = 0;
    int lowerCounter = 0;
    int charCounter[256] = { 0};//256 for all 8 bit values

    if ( NULL != ( input = fopen ( "input.txt", "r"))) {

        while ( ( letterTemp = fgetc ( input)) != EOF) {
            if ( islower ( letterTemp)) {//lower case
                lowerCounter++;
                charCounter[letterTemp]++;//increment count
            }
            if ( isupper ( letterTemp)) {//upper case
                upperCounter++;
                charCounter[tolower ( letterTemp)]++;//increment lower case count
            }
        }
        fclose ( input);
        letterMax = lowerCounter + upperCounter;

        if ( NULL != ( output = fopen ( "report.txt", "w"))) {
            fprintf ( output , "LETTER STATISTICS\n\n");
            fprintf ( output , "Category                How many in file    %% of all letters\n");
            fprintf ( output , "--------------------------------------------------------------\n");
            fprintf ( output , "Uppercase%31d%20.2f %%\nLowercase%31d%20.2f %%\n"
            , upperCounter
            , 100 * upperCounter / letterMax
            , lowerCounter
            , 100 * lowerCounter / letterMax);

            for (int i = 0; i < 256; i++) {
                if ( islower( i)) {//only print lower case values
                    fprintf ( output, "%c%39d%20.2f %%\n"
                    , i
                    , charCounter[i]
                    , 100 * charCounter[i] / letterMax);
                }
            }
            fclose ( output);
        }
        else {
            fprintf ( stderr, "could not open output file\n");
        }
    }
    else {
        fprintf ( stderr, "could not open input file\n");
    }
    return 0;
}

循环中的数组元素返回随机的大数，有时为负数

1 个答案: