C ++-cout标志仅适用于第一个输出

Question

When i pass some data, output is similiar to xxxxxxxx0x97 0x104 0x111 0x106 0x0 Why cout formatting is affecting only first output?

void Dumper::hex(const unsigned char * data, size_t len) {

        cout << endl;

        ios::fmtflags f(cout.flags());

        /*
        cout.fill('0');
        cout.width(2);
        */

        cout.fill('x');
        cout.width(10);

        for (int i = 0; i < len; i++) {
            cout << "0x" << ((long)(data[i]) & 255) << " ";
        }

        cout.flags(f);
        cout << endl;

    }

Answer 1

The width() is reset to 0 by any operation using it. The basic idea is that it is unlikely to apply to multiple fields. In particular, it unlikely to apply to a value and a separator. Thus, the width should be set up prior to each value.

For you specific use I'd use std::internal and std::showbase . Also make sure use unsigned values for your two digit hex values: otherwise you'll get the sign extended.