Question

我仍然坚持这个问题，我已经尝试了很多东西来获得它并且仍然没有成功......我想要做的就是将数字读入一个数组，将字母读入另一个数组。这样，我可以同时打电话给两本书的字母频率进行统计分析。我知道如何为字符串执行此操作，但我只想复制数字的输入，这些数字是它们所代表的字母前面的两个空格。代码通过命令提示符（argv [1]和argv [2]）读取两个文件，并将两者结合起来，取两者的均方根（RMS）作为输出。这是我到目前为止的代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <math.h>
#define NUM_LETTERS 26
int main( int argc, char *argv[] )
{
    FILE *fp, *fp2;
    int ch, i, counter;
    double frequencysquared[NUM_LETTERS], freqone[NUM_LETTERS], freqtwo[NUM_LETTERS], average;
    fp = fopen(argv[1], "r");
    fp2 = fopen(argv[2], "r");
    while((ch=fgetc(fp)) != EOF)
    {
        freqone[ch] = fgets(/*unsure*/, 22, fp); 
    }
    while((ch=fgetc(fp2)) != EOF)
    {
        freqtwo[ch] = fgets(/*unsure*/, 22, fp2);
    }
    while(i=0; i<NUM_LETTERS; i++)
    {
        average += pow(( freqone[i]-freqtwo[i] ), 2);
    }
    average/=NUM_LETTERS;
    average = sqrt(average);
    printf("RMS Frequency: %lf", average);
    fclose(fp);
    fclose(fp2);
    return 0;
}

文字档案：

A 0.08030130328545595500
B 0.01404566680148545800
C 0.02309245337888202900
D 0.04680329046987134100
E 0.12475974957130967000
F 0.02262448732647651800
G 0.02092142515718645500
H 0.06495870199587520900
I 0.06832638626586488900
J 0.00118328558965393900
K 0.00796545073487383260
L 0.03678714564106335500
M 0.02553256208071077300
N 0.07071134182580297900
O 0.07759211410349403600
P 0.01653257210855475400
Q 0.00111309068179311220
R 0.06200383063640040700
S 0.06269909448568859700
T 0.09005003894146078300
U 0.02792921679195900500
V 0.00865068674018190480
W 0.02356209073861756000
X 0.00115988728703366340
Y 0.02033479628434954300
Z 0.00035933107595423293

Answer 1

程序的粗略轮廓是正确的，但你有很多细节错误。例如，scanf可以负责获取输入。不需要乱拨字符串。

此外，您需要更好的错误检查。

这是一个开始：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define SIZE ('Z' - 'A' + 1)

void die(int err) {
  fprintf(stderr, "error %d\n", err);
  exit(err);
}

int main(int argc, char *argv[]) {
  char ch;
  double f, a[SIZE], b[SIZE];
  FILE *fa, *fb;

  if (argc != 3) die(1);
  fa = fopen(argv[1], "r");
  if (!fa) die(2);
  fb = fopen(argv[2], "r");
  if (!fb) die(3);
  for (int i = 0; i < SIZE; ++i)
    a[i] = b[i] = -1.0;
  while (fscanf(fa, " %c%lf", &ch, &f) == 2)
    if ('A' <= ch && ch <= 'Z') a[ch - 'A'] = f; else die(4);
  fclose(fa);
  while (fscanf(fb, " %c%lf", &ch, &f) == 2)
    if ('A' <= ch && ch <= 'Z') b[ch - 'A'] = f; else die(5);
  fclose(fb);
  double sum_d2 = 0;
  for (int i = 0; i < SIZE; ++i) {
    if (a[i] < 0 || b[i] < 0) die(100 + i);
    double d = a[i] - b[i];
    sum_d2 += d * d;
  }
  double rms = sqrt(sum_d2 / SIZE);
  printf("RMS Frequency: %f\n", rms);
  return 0;
}

Answer 2

一次只做一步。使代码尽可能简单。最好使用单独的函数来计算RMS，因为主函数中的所有内容都会变得无法控制。

以下是一些可帮助您完成任务的示例代码：

#include <stdio.h>
#include <stdlib.h>
#include <math.h>

#define NUMLETTERS 26

#define RMS_VALID 1
#define RMS_INVALID 0

double calc_rms(double X[], double Y[], int n, double *rms);
void read_file(double numbers[], char letters[], FILE *stream);

int 
main(int argc, char const *argv[]) {
    FILE *fp1, *fp2;
    double numbers1[NUMLETTERS], numbers2[NUMLETTERS];
    char letters1[NUMLETTERS], letters2[NUMLETTERS];
    double rms;

    fp1 = fopen(argv[1], "r");
    fp2 = fopen(argv[2], "r");
    if (fp1 == NULL || fp2 == NULL) {
        fprintf(stderr, "%s\n", "Error Reading a File");
        exit(EXIT_FAILURE);
    }

    read_file(numbers1, letters1, fp1);
    read_file(numbers2, letters2, fp2);

    if (calc_rms(numbers1, numbers2, NUMLETTERS, &rms) != RMS_INVALID) {
        printf("RMS Frequency: %f\n", rms);
    }

    return 0;
}

void
read_file(double numbers[], char letters[], FILE *stream) {
    double onedouble;
    char oneletter;
    int count = 0;

    while (fscanf(stream, " %c %lf", &oneletter, &onedouble) == 2) {
        numbers[count] = onedouble;
        letters[count] = oneletter;
        count++;
    }
}

double
calc_rms(double X[], double Y[], int n, double *rms) {
    int i;
    double sum = 0.0, diff;

    if (n <= 0) {
        return RMS_INVALID; // just for precaution
    }

    for (i = 0; i < n; i++) {
        diff = (X[i] - Y[i]);
        sum += diff * diff;
    }
    *rms = sqrt(sum/n);
    return RMS_VALID;
}

从文本文件

2 个答案: