二进制转换成八进制、十进制、十六进制

 1、二进制转换成十进制

#include 

int main()
{
    int remainder, number = 0, decimal_number = 0, temp = 1;
    printf("\n Enter any binary number= ");
    scanf("%d", &number);

    while (number > 0)
    {
        remainder = number % 10;
        number = number / 10;
        decimal_number += remainder * temp;
        temp = temp * 2;  
    }

    printf("%d\n", decimal_number);
}

测试结果：

 Enter any binary number= 1010
10 

 2、二进制转换成八进制

#include 

int three_digits(int n)
{
    int r, d = 0, p = 1;

    for (int i = 0; i < 3; i++)
    {
        r = n % 10;
        d += r * p;
        p *= 10;
        n /= 10;
    }
    return d;
}

int main(void)
{
    int binary_num, d = 0, base = 1, remainder, td, res = 0, ord = 1;

    printf("Enter the binary no: ");
    scanf("%d", &binary_num);

    while (binary_num > 0)
    {
        if (binary_num > 111) 
            td = three_digits(binary_num);

        else
            td = binary_num;

        binary_num /= 1000;

        d = 0, base = 1;

        while (td > 0)
        {
            remainder = td % 10;
            td /= 10;
            d += (base * remainder);
            base *= 2;
        }

        res += d * ord;  
        ord *= 10;
    }

    printf("\nOctal equivalent is: %d", res);
    return 0;
}

 测试结果：

Enter the binary no: 1010

Octal equivalent is: 12

3、二进制转换成十六进制

#include 

int main()
{
    long int binary, hexa = 0, i = 1, remainder;

    printf("Enter the binary number: ");
    scanf("%ld", &binary);
    while (binary != 0)
    {
        remainder = binary % 10;
        hexa = hexa + remainder * i;
        i = i * 2;
        binary = binary / 10;
    }
    printf("The equivalent hexadecimal value: %lX", hexa);
    return 0;
}

测试结果：

Enter the binary number: 1010
The equivalent hexadecimal value: A

 4、字符串转换成整型

#include 
#include 
#include 
#include 

int c_atoi(const char *str)
{
    int i;
    int sign;
    long value;
    long prev;

    i = 0;
    sign = 1;
    value = 0;

    /* 跳过空格 */
    while (((str[i] <= 13 && str[i] >= 9) || str[i] == 32) && str[i] != '\0')
        i++;

    if (str[i] == '-')
	{
        sign = -1;
	i++;
	}
    else if (str[i] == '+')
	{
        sign = 1;
	i++;
	}

    while (str[i] >= 48 && str[i] <= 57 && str[i] != '\0')
    {
        prev = value;
        value = value * 10 + sign * (str[i] - '0');

        if (sign == 1 && prev > value)
            return (-1);
        else if (sign == -1 && prev < value)
            return (0);
        i++;
    }
    return (value);
}

void test_c_atoi()
{
    printf("<<<< TEST FUNCTION >>>>\n");
    assert(c_atoi("123") == atoi("123"));
    assert(c_atoi("-123") == atoi("-123"));
    assert(c_atoi("") == atoi(""));
    assert(c_atoi("-h23") == atoi("-h23"));
    assert(c_atoi("         23") == atoi("         23"));
    assert(c_atoi("999999999") == atoi("999999999"));
    printf("<<<< TEST DONE >>>>\n");
}


int main(int argc, char **argv)
{
    test_c_atoi();

    if (argc == 2)
    {
        printf("Your number + 5 is %d\n", c_atoi(argv[1]) + 5);
        return (0);
    }
    printf("wrong number of parmeters\n");
    return (1);
}

测试结果：

./c_atoi "123"
<<<< TEST FUNCTION >>>>
<<<< TEST DONE >>>>
Your number + 5 is 128

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