无法在C中显示文件的内容

Question

我试图读取一个名为的文件的内容,pp.txt并在命令行上显示它的内容.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;
}

编译后我得到了File opened successfully!File read failed.我的内容pp.txt是34.5.谁能告诉我哪里出错了？

Answer 1

问题是你正在执行两次你的一些功能.这里:

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

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

和这里:

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

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

改变他们

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;
}

别忘了定义int r;

您收到错误是因为您的第一个fscanf()调用读取了数字并将文件指针移到了它之外.现在第二个呼叫找不到号码而失败.