问题
I have a simple question about using fgets() with char* string.
....
char *temp;
FILE fp=fopen("test.txt", "r");
fgets(temp, 500, fp);
printf("%s", temp);
....
This code didn't work well.
But after I modified char *temp to char temp[100];, the code worked well as I intended.
What is the difference between those two?
When I googled it, some said that memory must be allocated to char * using malloc()...
But I couldn't understand it.
回答1:
char *temp is only a pointer. At begin it doesn't points to anything, possibly it has a random value.
fgets() reads 500 bytes from fp to the memory addresse, where this temp pointer points! So, it can overwrite things, it can make segmentation faults, and only with a very low chance will be work relativale normally.
But char temp[500] is a 500 bytes long array. That means, that the compiler does the allocation on the beginning of your process (or at the calling of your function). Thus this 500 bytes will be a useable 500 bytes, but it has a price: you can't reallocate, resize, free, etc. this.
What the google wants from you, is this:
char *temp = (char*)malloc(500);
And a
free(temp);
after you don't need this any more.
回答2:
char *temp is uninitialized, that is, it isn't pointing to valid memory. Either make it an array (char temp[]) or use malloc to assign memory for it.
回答3:
When we write
char *temp ;
it means temp is an uninitialized pointer to char i.e. currently it does not contain any address in it .
While using fgets you have to pass a string in which the bytes read from file pointer is to be copied . link
since the temp is uninitialized , the fgets looks like this
fgets(<no string> , 500 , fp ) ;
which is invalid .
Hence , we should give initialized string which can be formed as :
1) char *temp = malloc(sizeof(500)) ;
or
2) char temp[500] ;
Hence if we pass initialized string to fgets , it would look like
fgets( < some string > , 500 , fp) ;
来源：https://stackoverflow.com/questions/20198180/using-fgets-with-char-type