istringstream decimal integer input to 8-bit type

Question

This:

#include <iostream>
#include <sstream>
#include <inttypes.h>

using namespace std;

int main (void) {
    istringstream iss("123 42");
    int8_t x;
    while (iss >> x) {
        cout << x << endl;
    }
    return 0;
}  

Produces:

1
2
3
4
2

But I want:

123
42

Casting iss >> (int)x (I initially tried this with a char) gives me "error: invalid operands to binary expression ('istringstream' (aka 'basic_istringstream') and 'int')" (clang) or "error: ambiguous overload for ‘operator>>’" (g++).

Is there a way to read the value as a number directly into an 8-bit type, or do I have to use an intermediary store?

Answer 1

There is no built-in 8-bit type; you're using an alias for signed char and IOStreams will always extract a single ASCII letter when you do formatted input into any kind of char.

So, yes, use an intermediary store, or wrap int8_t in a new class that provides its own overloads for formatted I/O (which I'd consider overkill unless you have strict memory and/or performance requirements).

(Your attempt of iss >> (int)x is very confused; conversions are used on expressions you're about to take the value of, not for lvalues naming objects that you want to set the value of.)

Answer 2

You have to use an intermediate type or do the parsing yourself. All char-types (char, signed char and unsigned char) are treated as text elements, not integers. int8_t is probably just a typedef for one of them, which is why your code fails.

Note:

• The output will suffer from the same issues.
• Don't use C-style casts, they almost only cause errors.
• Checking for EOF before an input operation is useless, you need to check for failure afterwards instead.