printf（）参数传递

//详见c primer plus 5th 中文 p75 参数传递
#include <stdio.h>


int main(void)
{
    float n1 = 3.0;
    double n2 = 3.0;
    long n3 = 2000000000;
    long n4 = 1234567890;
    long n5, n6;


    printf("%.1e %.1e %.1e %.1e\n", n1, n2, n3, n4);
    printf("%ld %ld\n", n3, n4);
    printf("%ld %ld %ld %ld %ld %ld\n", n1, n2, n3, n4, n5, n6);


    return 0;
}


/*
在codeblocks 13.12 输出结果：
3.0e+000 3.0e+000 3.1e+046 2.4e+261
2000000000 1234567890
0 1074266112 0 1074266112 2000000000 1234567890
*/