从整数中提取数字时出错

Question

I have a simple program which gives the wrong output, the expected output is the digits of the number.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>


using namespace std;

int main()
{
   int n = 125521;
   int d = floor(log10(n));
   printf("%d Digits\n",d+1);
   int t =0;
   while(floor(log10(n))-t)
   { printf("%d-----%d\n",(n/(int)pow(10,floor(log10(n))-t)%10),t); t++;}
   return 0;
}

This gives the output

6 Digits
1-----0
2-----1
5-----2
7-----3
2-----4

Strange output. Why does 7 come ?

I know how to get the digits by other ways but I want this solution to work.

Now as suggested in the answers I get rid of the bug in the while loop (>=0) and I get the output:

Answer 1

Put proper spaces (n/(int)pow(10,floor(log10(n))-t)%10) in this line.

So many redundant call of floor(log10(n)) .

while loop will loop less than 1 from digit numbers.

I have no idea, why you print t besides digits, say:

1-----0
2-----1
     ^^^

Can't reproduce what you said.

Are you looking for this:

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstdlib>

using namespace std;

int main() {
    int n = 550052111;
    int d = floor(log10(n));
    printf("%d Digits\n", d + 1);
    int t = 0;
    int power, divition, mod;
    int digitCount[10] = {0};
    while (d +1 - t) {
        power = (int) pow(10, d - t);
        divition = n / power;
        mod = divition % 10;
        digitCount[mod]++;
        t++;
    }

    for (t= 0; t < 10; t++) {
        if(digitCount[t]) {
            printf("%d-----%d\n", t, digitCount[t]);
        }
    }
    return 0;
}

Output:

9 Digits
0-----2
1-----3
2-----1
5-----3

Caution: input n must not overflow Integer . To find the digit counts in a number, you do not need floor , pow or log10 , only / and % is sufficient.

Answer 2

There is a bug in your while loop.

while(floor(log10(n))-t)

Should be

while(0 <= floor(log10(n))-t)

The index in your case should start from 0 rather than 1.

Fix that bug and you should get the proper output as follows:

6 Digits
1-----0
2-----1
5-----2
5-----3
2-----4
1-----5

Answer 3

To find out why you are getting the 7 instead of the 5, you need to do some serious diagnostic printing — maybe using code something like this:

#include <cstdio>
#include <cmath>

using namespace std;

int main()
{
    int n = 125521;
    int d = floor(log10(n));
    printf("n = %d: %d Digits\n", n, d + 1);
    int t = 0;
    while (floor(log10(n)) >= t)
    {
        printf("t = %d: n = %d; L = log10(n) = %f; F = floor(L) = %f\n",
                t, n, log10(n), floor(log10(n)));
        printf("       T = F-t = %f;", floor(log10(n)) - t);
        printf(" P = pow(10,T)= %f\n", pow(10, floor(log10(n)) - t));
        printf("       I = (int)P = %d; D = n/I = %d; M = D %% 10 = %d\n",
               (int)pow(10, floor(log10(n)) - t),
               n / (int)pow(10, floor(log10(n)) - t),
               n / (int)pow(10, floor(log10(n)) - t) % 10);
        printf("%d-----%d\n", (n / (int)pow(10, floor(log10(n)) - t) % 10), t);
        t++;
    }
    return 0;
}

I've modified the loop condition; the old-style Fortran II arithmetic if condition really isn't good style in C. It also prints the last digit. It would be better if the loop were written for (int t = 0; t <= d; t++) , but I've not made that change.

On a Mac running OS X 10.10.3, using GCC 5.1.0 compiling a 64-bit program, the result is:

n = 125521: 6 Digits
t = 0: n = 125521; L = log10(n) = 5.098716; F = floor(L) = 5.000000
       T = F-t = 5.000000; P = pow(10,T)= 100000.000000
       I = (int)P = 100000; D = n/I = 1; M = D % 10 = 1
1-----0
t = 1: n = 125521; L = log10(n) = 5.098716; F = floor(L) = 5.000000
       T = F-t = 4.000000; P = pow(10,T)= 10000.000000
       I = (int)P = 10000; D = n/I = 12; M = D % 10 = 2
2-----1
t = 2: n = 125521; L = log10(n) = 5.098716; F = floor(L) = 5.000000
       T = F-t = 3.000000; P = pow(10,T)= 1000.000000
       I = (int)P = 1000; D = n/I = 125; M = D % 10 = 5
5-----2
t = 3: n = 125521; L = log10(n) = 5.098716; F = floor(L) = 5.000000
       T = F-t = 2.000000; P = pow(10,T)= 100.000000
       I = (int)P = 100; D = n/I = 1255; M = D % 10 = 5
5-----3
t = 4: n = 125521; L = log10(n) = 5.098716; F = floor(L) = 5.000000
       T = F-t = 1.000000; P = pow(10,T)= 10.000000
       I = (int)P = 10; D = n/I = 12552; M = D % 10 = 2
2-----4
t = 5: n = 125521; L = log10(n) = 5.098716; F = floor(L) = 5.000000
       T = F-t = 0.000000; P = pow(10,T)= 1.000000
       I = (int)P = 1; D = n/I = 125521; M = D % 10 = 1
1-----5

You should run either this program or something very similar and show the results of the intermediate calculations. We can then, maybe, begin to understand why you see a 7 instead of a 5.