How Does C++'s atoi work?

Question

So I have the following code:

void Start(int &year, string &mon, char &nyd)
{
    printf("%s", mon);
    int month= atoi(mon.c_str());
    printf("%i", month);
}

When the incoming parameter is "03" (the first printf shows 03), I got 0 for month.

However, if I add this line

mon = "03";

I got 3, which is correct, for month.

Why......????

EDIT: I figured it out. You guys were right. DON'T use scanf for string input.

With an addition of `.data()` or `.c_str()`. Even better, stick with something better than `atoi`, such as `stoi`, which actually has error checking and more relevant to that line, `std::cout`, which is type safe for printing. — chris, Jan 29 '13 at 03:46
Nevertheless, the result of the atoi function, which is month, is not correct... — user1447343, Jan 29 '13 at 03:47

score 3 · Answer 1 · edited Jan 29 '13 at 04:07

3

You can't print std::string with %s in printf function, try this:

void Start(int &year, const std::string &mon, char &nyd)
{
    std::cout << mon << std::endl;
    int month= atoi(mon.c_str());
    std::cout << month << std::endl;
}

Or

void Start(int &year, string &mon, char &nyd)
{
    printf("%s\n", mon.c_str());
    int month= atoi(mon.c_str());
    printf("%i\n", month);
}

But std::cout is preferred over C printf function.

Also don't use scanf with std::string, use std::cin instead of scanf, std::cout instead of printf.

edited Jan 29 '13 at 04:07

Jonathan Leffler

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answered Jan 29 '13 at 03:47

billz

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Yeah, but the problem is not about the printf; it's about atoi. I still get 0 when mon = "03" after changing the printf function. – user1447343 Jan 29 '13 at 03:50
printf("Month: ");scanf("%s", month);Start(year, month, nyd); – user1447343 Jan 29 '13 at 03:54
3

@user1447343 Don't use scanf with std::string – Rapptz Jan 29 '13 at 03:55

How Does C++'s atoi work?

1 Answers1