Using fgets() with char* type

Question

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\", te         


        

Answer 1

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) ;

Answer 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.

Answer 3

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.