c - C 无法显示文件内容

Question

我正在尝试读取名为 pp.txt 的文件的内容并在命令行上显示其内容。我的代码是:

#include<stdio.h>
#include<stdlib.h>
int main()
{

FILE *f;
float x;


f=fopen("pp.txt", "r");

if((f = fopen("pp.txt", "r")) == NULL)
{
fprintf(stderr, "Sorry! Can't read %s\n", "TEST1.txt");
}

else
{
printf("File opened successfully!\n");
}

fscanf(f, " %f", &x);

if (fscanf(f, " %f ", &x) != 1) {
fprintf(stderr, "File read failed\n");
return EXIT_FAILURE;
}

else
{
printf("The contents of file are: %f \n", x);
}


fclose(f);

return 0;
}

编译后，我得到文件打开成功!文件读取失败。我的 pp.txt 内容是 34.5。谁能告诉我哪里出错了？

Answer 1

问题是您执行了某些函数两次。这里:

f=fopen("pp.txt", "r");

if((f = fopen("pp.txt", "r")) == NULL)
{
fprintf(stderr, "Sorry! Can't read %s\n", "TEST1.txt");
}

这里:

fscanf(f, " %f", &x);

if (fscanf(f, " %f ", &x) != 1) {
fprintf(stderr, "File read failed\n");
return EXIT_FAILURE;
}

将它们更改为

f=fopen("pp.txt", "r");

if(f == NULL)
{
  fprintf(stderr, "Sorry! Can't read %s\n", "TEST1.txt");
  return EXIT_FAILURE;
}

和

r = fscanf(f, " %f", &x);

if (r != 1) 
{
  fclose(f); // If fscanf() fails the filepointer is still valid and needs to be closed
  fprintf(stderr, "File read failed\n");
  return EXIT_FAILURE;
}

不要忘记定义 int r;

您收到错误是因为您的第一个 fscanf() 调用读取了该数字并将文件指针移到了该数字之外。现在第二次调用找不到号码并失败。