void print_each_digit(int x)
{
    if(x >= 10)
       print_each_digit(x / 10);

    int digit = x % 10;

    std::cout << digit <<  
 ;
}

因此,它要加以扼杀,然后就特性进行 it弄。 您可使用<代码>std:ostringstream,例如:

int iNums = 12476;
std::ostringstream os;

os << iNums;
std::string digits = os.str();

Btw the generally used term (for what you call "number") is "digit" - please use it, as it makes the title of your post much more understandable :-)

这里是一种比较通用的、可再生的解决办法,可产生数字媒介:

void collect_digits(std::vector<int>& digits, unsigned long num) {
    if (num > 9) {
        collect_digits(digits, num / 10);
    }
    digits.push_back(num % 10);
}

由于数字相对较少,重新入侵受到严格限制。

www.un.org/Depts/DGACM/index_spanish.htm 这里是采取这一行动的途径,但你会按相反顺序行事。

int num;
short temp = 0;
cin>>num;

while(num!=0){
    temp = num%10;
    
    //here you will get its element one by one but in reverse order
    //you can perform your action here.
    
    num /= 10;
}

我没有考验它只是写一下我的头脑。 任何辛加税错误的借口

http://ideone.com/flbAQ“rel=“nofollow noreferer”>online ideone demo。

vector <int> v; 

int i = ....
while(i != 0 ){
    cout << i%10 << " - "; // reverse order
    v.push_back(i%10); 
    i = i/10;
}

cout << endl;

for(int i=v.size()-1; i>=0; i--){
   cout << v[i] << " - "; // linear
}

在“拥有”职位上取得数位数(从职位1起算为最不重要的数字数字):

digit = (int)(number/pow(10,(pos-1))) % 10;

例:编号=57820 ->pos = 4 ->数 = 7

依次获得数字:

int num_digits = floor( log10(abs(number?number:1)) + 1 );
for(; num_digits; num_digits--, number/=10) {
    std::cout << number % 10 << " ";
}

例:编号=57820 ->输出:0 2 8 7 5

你可以履行这一职能:

void printDigits(int number) {
    if (number < 0) { // Handling negative number
        printf( - );
        number *= -1;
    }
    if (number == 0) { // Handling zero
        printf( 0 );
    }
    while (number > 0) { // Printing the number
        printf("%d-", number % 10);
        number /= 10;
    }
}

从D.Shawley的答复中可以得出以下结果:

void stream_digits(std::ostream& output, int num, const std::string& delimiter = "")
{
    if (num) {
        stream_digits(output, num/10, delimiter);
        output << static_cast<char>( 0  + (num % 10)) << delimiter;
    }
}

void splitDigits()
{
    int num = 12476;    
    stream_digits(std::cout, num, "-");    
    std::cout << std::endl;
}

我不知道这是否更快或更慢或毫无价值,但这将是一种选择:

int iNums = 12476;
string numString;
stringstream ss;
ss << iNums;
numString = ss.str();
for (int i = 0; i < numString.length(); i++) {
    int myInt = static_cast<int>(numString[i] -  0 ); //  0  = 48
    printf("%i-", myInt);
}

我指出,“iNums”是指有可能成为用户投入,如果用户投入首先是一个扼杀物,那么你就不需要穿过将暗中转换为扼子的通道。

(to_string could be used in c++11)

我知道这是一个老的职位,但所有这些答案对我来说都是不可接受的。

我的目的是把一些人员放在屏幕上,从而提供功能名称。

void RenderNumber(int to_print)
{
    if (to_print < 0)
    {
        RenderMinusSign()
        RenderNumber(-to_print);
    }
    else
    {
        int digits = 1; // Assume if 0 is entered we want to print 0 (i.e. minimum of 1 digit)
        int max = 10;

        while (to_print >= max) // find how many digits the number is 
        {
            max *= 10;
            digits ++;
        }

        for (int i = 0; i < digits; i++) // loop through each digit
        {
            max /= 10;
            int num = to_print / max; // isolate first digit
            to_print -= num * max; // subtract first digit from number
            RenderDigit(num);
        }
    }
}

根据@Abyx的答复,但使用div 因此,每个数字只有1个司。

#include <cstdlib>
#include <iostream>

void print_each_digit(int x)
{
    div_t q = div(x, 10);

    if (q.quot)
        print_each_digit(q.quot);

    std::cout << q.rem <<  - ;
}

int main()
{
    print_each_digit(12476);
    std::cout << std::endl;
    return 0;
}

产出:

1-2-4-7-6-

N.B. Only works for non-negative ints.

我的解决办法:

void getSumDigits(int n) {
    std::vector<int> int_to_vec;
    while(n>0)
    {
        int_to_vec.push_back(n%10);
        n=n/10;
    }

    int sum;

    for(int i=0;i<int_to_vec.size();i++)
    {
        sum+=int_to_vec.at(i);

    }
    std::cout << sum <<    ;
}

所用的答案一是这一简单功能:

int getDigit(int n, int position) {

    return (n%(int)pow(10, position) - (n % (int)pow(10, position-1))) / (int)pow(10, position-1);

}

希望有人会认为这有助益!

在线C++汇编者,在线管理C++方案

#include <iostream>
#include <cmath>

int main() {
    int iNums = 123458;

//t iNumsSize = 5;

int iNumsSize = trunc(log10(iNums)) + 1; // Find length of int value
for (int i=iNumsSize-1; i>=0; i--) {
    int y = pow(10, i); 
// The pow() function returns the result of the first argument raised to 

the power of the second argument.
    int z = iNums/y;
    int x2 = iNums / (y * 10);
    printf("%d ",z - x2*10 ); // Print Values
}

    return 0;
}

You can do it using a while loop and the modulo operators. It just gives the digits in the revese order.

int main() {
    int iNums = 12476;
    int iNum = 0;
    while(iNums > 0) {
        iNum = iNums % 10;
        cout << iNum;
        iNums = iNums / 10;
    }
}

int a;
cout << "Enter a number: ";
cin >> a;

while (a > 0) {
    cout << a % 10 << endl;
    a = a / 10;             
}