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Editorial Team
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Asked: 2026-05-20T15:18:37+00:00

#include <stdio.h> #include <string> int main(void) { printf(%u\n, std::string(\n).find_first_not_of( \t\n\v\f\r, 0, 1)); } The

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#include <stdio.h>
#include <string>

int main(void)
{
    printf("%u\n", std::string("\n").find_first_not_of(" \t\n\v\f\r", 0, 1));
}

The following program prints 0, not std::string::npos as I expected. Why?

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1 Answer

  1. Editorial Team

    Your call matches:

    size_t find_first_not_of ( const char* s, size_t pos, size_t n ) const;

    n is the number of chars in s, and you’re passing 1. So, you’re searching for the first char that is not space. The rest of your " \t\n\v\f\r" string is ignored.

    Likely you simply want:

    find_first_not_of(" \t\n\v\f\r")
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