In c: Given the following code

int a;
char word[]=<some string>

Is there a difference between?

a = atoi(word) 
a = atof(word)

CodePudding user response：

atoi returns integer type, atof returns double.

So in the atof scenario you are additionally converting the temporary double into your int

CodePudding user response：

Yes, you can.

But the compiler will emit warnings. See below:

It tells you what will happen.

All double or float values will be truncated (not rounded).

See the below code:

#include <iostream>
#include <cstdlib>

int main() {
 
    int i1{};
    int i2{};
    int i3{};
    char s1[] = "42";
    char s2[] = "42.1";
    char s3[] = "42.9";

    i1 = std::atof(s1);
    i2 = std::atof(s2);
    i3 = std::atof(s3);

    std::cout << s1 << "\t--> " << i1 << '\n'
        << s2 << "\t--> " << i2 << '\n'
        << s3 << "\t--> " << i3 << '\n';
}

If you want to get rid of the warnings and be more clean, you need to add a cast statement. Like below:

#include <iostream>
#include <cstdlib>

int main() {

    int i1{};
    int i2{};
    int i3{};
    char s1[] = "42";
    char s2[] = "42.1";
    char s3[] = "42.9";

    i1 = static_cast<int>(std::atof(s1));
    i2 = static_cast<int>(std::atof(s2));
    i3 = static_cast<int>(std::atof(s3));

    std::cout << s1 << "\t--> " << i1 << '\n'
        << s2 << "\t--> " << i2 << '\n'
        << s3 << "\t--> " << i3 << '\n';
}

There will be no compiler warning:

Program output will be: