using a part of code like this:
fscanf(f1, "%d", &n);
while(!feof(f1)){
...
fscanf(f1, "%d", &n);}
it misses the last line of the file (meeting EOF). How should I solve?
the only solution i found is:
if (fscanf(f1, "%d", &n)!=EOF){
rewind(f1);
do{
...
fscanf(f1, "%d", &n);
}while(!feof(f1));
}
您将 fscanf 放在循环的末尾。fscanf 读取直到确定数字已完成。如果输入文件的最后一个字符是数字（与空格或换行相对），则在解析最后一行时（有些人会将不以换行结尾的行称为“不完整的最后一行”），fscanf 会遇到 EOF 试图找到数字的结尾，所以 feof 是真的，因为 EOF 已经被命中了。
您不应该检查 feof，而应该检查 fscanf 的返回码。它会告诉你是否有一些数字可以被解析为数字。
假设您的文件包含"11\n23"
f = fopen(...);
result = fscanf(f, "%d", &i);
// result == 1, because one variable has been read
// i == 11, because that's the first number
// file pointer is past the '\n', because '\n'
// had to be read to find out the number is not
// something like 110
// The '\n' itself has been put back using ungetc
// feof(f) == 0, because nothing tried to read past EOF
result = fscanf(f, "%d", &i);
// result == 1, because one variable has been read by this call
// i == 23 (obviously)
// file pointer is at EOF (it can't go further)
// feof(f) == 1, because fscanf tried to read past
// the '3' to check whether there were extra
// characters.
// (Your loop terminates here, because feof(f) is true
result = fscanf(f, "%d", &i);
// result == EOF (error before first variable)
// i is likely unchanged. I am unsure whether this
// is guaranteed by the language definition
// file pointer unchanged
// feof(f) still true
// You should terminate processing *NOW*, because
// return is no longer one.