Method sscanf() ambiguous behavior

Question

I am trying to understand how sscanf() works. I ran some examples from this page: http://docs.roxen.com/pike/7.0/tutorial/strings/sscanf.xml and they don't work on my platform. I can't understand why. For instance: "sscanf("4711bar", "%d%s", a, b);" makes the program exit with an error...

Here is one of the examples that work: "sscanf("foo", "f%s", a);". Does anybody know why? Do they work on your platforms? Thank you.

This is my code:

 int main(void){

   char *b = (char*)malloc(sizeof(char)*100);       
   int a = 0;

   sscanf("4711bar", "%d%s", a, b);
   printf("%d", a);
   printf("%s", b);
 }

The documentation you are looking at **is not for C or C++ but for an entirely different language**. — n. m. could be an AI, Nov 09 '13 at 20:12
Also, [don't cast the return value of `malloc()`](http://stackoverflow.com/questions/605845/do-i-cast-the-result-of-malloc/605858#605858). — , Nov 09 '13 at 20:15
@CarmenCojocaru Because it is an error to do so, and you don't want to have errors in your code (care to read the answer behind the link?) — , Nov 09 '13 at 20:22
@H2CO3 This comes from the same article: "True. However, in C++ the cast is required, so if you want your code to work in **both**, you'll have to compromise. But in pure C, don't do the cast for the reasons you stated. " — Carmen Cojocaru, Nov 09 '13 at 20:25
@CarmenCojocaru But no sane programmer uses `malloc()` in C++. In C++, one uses `vector` or at most `operator new[]`. If you *need* the cast, that indicates that you are doing something terribly wrong. — , Nov 09 '13 at 20:26

wkordalski · Accepted Answer · 2015-06-17T19:16:31.360

1

You should write: sscanf("4711bar", "%d%s", &a, b); The & makes &a is a pointer to a and sscanf requires pointers.

The second example works, because a is an array. And array a of type char[100] is implictly converted to pointer of type char * pointing to the array's first element (a[0]). So a is the same as &(a[0])

int a;
char b[100];
sscanf("4711bar", "%d%s", &a, b);

-

char a[100];
sscanf("foo", "%s", a);

edited Jun 17 '15 at 19:16

answered Nov 09 '13 at 20:11

wkordalski

938
8
9

Why it doesn't work with a pointer to char, like char *b and it does with an array of chars? – Carmen Cojocaru Nov 09 '13 at 20:22
"And name of array is pointer to its first element." - it isn't, but it is **implicitly converted into** a pointer to its first element. – Nov 09 '13 at 20:24
@CarmenCojocaru Because the uninitialized pointer doesn't point to any valid memory. – Nov 09 '13 at 20:24
@Zifre No problem, it's a mistake easy to make. (At least you didn't insist! How many people do...) – Nov 09 '13 at 20:29

score 0 · Answer 2 · answered Nov 09 '13 at 20:13

You should pass valid pointers to sscanf, &a for integer and b for string:

sscanf("4711bar", "%d%s", &a, b);
                          ^

Moreover, your code is C-ish not C++. Using C++ it should looks something like:

std::stringstream s;
s << "4711bar";

int a;
std::string b;

s >> a >> b;

Method sscanf() ambiguous behavior

2 Answers2