Question

我在文件data1.txt中有2行（如下所示）：

Da    KOL    -1.19503   5.27557163                      
MaB   KOL    -1.19503   5.27557163

我不确定如何使用fgets提取特定的单词或数字，因此我使用fscanf扫描集合的每个组件并将它们打印到另一个文件。代码是：

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


main()
{
 FILE *fpt1, *fpt2;
    fpt1=fopen("data1.txt","r");
    fpt2=fopen("data2.txt","w");

    int i;
    double  ep, si;
    char *sto1, *sto2;

    for(i=0;i<2;i++)
     {  
       fscanf(fpt1,"%s\n",sto1);
       fscanf(fpt1,"%s\n",sto2);
       fscanf(fpt1,"%lf\n",&ep);
       fscanf(fpt1,"%lf\n",&si);
       fprintf(fpt2,"%s %s %2.8lf %2.8lf\n",sto1,sto2,ep,si);
     }
 fclose(fpt1);
 fclose(fpt2);  
}

但我在文件data2.txt中获取此输出：

Da (null) 0.00000000 0.00000000
KOL (null) -1.19503000 5.27557163

虽然，所需的输出是在一行的每个组件之间有一个空格的输出，如：

Da KOL -1.19503 5.27557163                      
MaB KOL -1.19503 5.27557163

有人可以帮我解决这个问题吗？

Answer 1

#include <stdio.h>

//A good habit is not to use "main()"
int main(int argc, char *argv[])
{
    FILE *fpt1, *fpt2;
    fpt1=fopen("data1.txt","r");
    fpt2=fopen("data2.txt","w");

    int i;
    double ep, si;
    char sto1[100], sto2[100];  //here neeed array

    for(i=0;i<2;i++)
    {  
        fscanf(fpt1,"%s",sto1); //without '\n'
        fscanf(fpt1,"%s",sto2);
        fscanf(fpt1,"%lf",&ep);
        fscanf(fpt1,"%lf",&si);
        fprintf(fpt2,"%s %s %2.8lf %2.8lf\n",sto1,sto2,ep,si);
    }
    fclose(fpt1);
    fclose(fpt2);       

    return 0;
}

Answer 2

你的想法是正确的，但我已经修改/纠正了你的代码。我使用sscanf代替fscanf。这是代码，

#include<stdio.h>
#define MAX 256

int main(){

   FILE *pfile1 =NULL,*pfile2 = NULL;
   char *sto1=NULL,*sto2=NULL,line[MAX];
   double ep,si;

   pfile1 = fopen("dS.txt","r");
   pfile2 = fopen("dR.txt","w");

   if(pfile1 != NULL || pfile2 != NULL){
      while(fgets(line,255,pfile1)!=NULL){
        sscanf(line,"%s %s %Lf %Lf",&sto1,&sto2,&ep,&si);
        fprintf(pfile2,"%s %s %2.5Lf %2.8Lf\n",&sto1,&sto2,ep,si);
      }
      fclose(pfile1);
      fclose(pfile2);
  }
  return 0;
}

Answer 3

您使用了具有自动存储时间sto1和sto2的未初始化变量的值，并调用了未定义的行为。

在使用它们之前，指定指向它们的某些有效缓冲区的指针。

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

int main(void)
{
    FILE *fpt1, *fpt2;
    fpt1=fopen("data1.txt","r");
    fpt2=fopen("data2.txt","w");

    int i;
    double  ep, si;
    char *sto1, *sto2;

    /* allocate enough size to store data */
    sto1 = malloc(1000000);
    sto2 = malloc(1000000);

    for(i=0;i<2;i++)
    {
        fscanf(fpt1,"%s\n",sto1);
        fscanf(fpt1,"%s\n",sto2);
        fscanf(fpt1,"%lf\n",&ep);
        fscanf(fpt1,"%lf\n",&si);
        fprintf(fpt2,"%s %s %2.8lf %2.8lf\n",sto1,sto2,ep,si);
    }
    fclose(fpt1);
    fclose(fpt2);
    free(sto1);
    free(sto2);
}

为fopen()，malloc()和fscanf()添加错误检查会使此代码更好。

替代方法是使用静态分配的数组，而不是通过内存管理函数动态分配缓冲区。

#include<stdio.h>

int main(void)
{
    FILE *fpt1, *fpt2;
    fpt1=fopen("data1.txt","r");
    fpt2=fopen("data2.txt","w");

    int i;
    double  ep, si;
    char sto1[1000], sto2[1000]; /* allocating too big array as automatic local variable may cause stack overflow */

    for(i=0;i<2;i++)
    {
        fscanf(fpt1,"%s\n",sto1);
        fscanf(fpt1,"%s\n",sto2);
        fscanf(fpt1,"%lf\n",&ep);
        fscanf(fpt1,"%lf\n",&si);
        fprintf(fpt2,"%s %s %2.8lf %2.8lf\n",sto1,sto2,ep,si);
    }
    fclose(fpt1);
    fclose(fpt2);
}

您可以使用静态数组

扫描每个单词或数字并使用C打印到文件

3 个答案: