How get first three digits of an integer in C++

[Balakumar]

Hi

I want to know how to get the first three digits of a four digit number in C++.Any suggestion?

[helios]

Which first three digits?

[webJose]

If the number is 1234 and you want 123, then:

int value = 1234;
int newVal = value / 10;

If you instead want 234, then:

int value = 1234;
int newVal = value - value / 1000 * 1000;

At this point newVal can be printed or whatever.

[Gregor]

webjose what are you doing for 234, makes no sense

for 234:

int newVal = value%1000;

general:

get rid of last n digits in number a

a/10^n

get rid of first n digits in number a

a%10^n

[helios]

Actually, it does: x/y*z == (x/y)*z => 1234-1234/1000*1000 == 1234-1*1000 == 1234-1000 == 234

I still like the modulo version better, though.

10*e^n

I don't get what this is supposed to mean.

[Gregor]

exp(n)

should just be e^n as I check it now

[helios]

wat

1234/10*e^1 ~= 1234/27.183 ~= 45.4
1234/10^1 == 123 (integral division)

[Gregor]

dunno what got into me.

[webJose]

Gregor:

I didn't know the modulus operator in C++. :S

As stated by helios, mine work too.

[Alan]

truncation:

stringstream ss;
short number = 1234;
ss<< number;
short firstThreeNumbers = atoi(ss.str().substr(0, 3).c_str());
cout << firstThreeNumbers << '\n';

division of integer:

short number = 1234;
short firstThreeNumbers = number/10; // truncates remainder... floor()
cout << firstThreeNumbers << '\n';

[Alan]

@Gregor

get rid of first n digits in number a
a%10^n

I thought that a modulus returns:

((a/b)-floor(a/b))!=0

could you walk me through that

[Gregor]

In computing, the modulo operation finds the remainder of division of one number by another.

http://en.wikipedia.org/wiki/Modulo_operation

As simple as that.

[Alan]

In computing, the modulo operation finds the remainder of division of one number by another.

Why is is in C++ either 0 or 1 is returned?

int n1 = 5;
int n2 = 4;
int imod1 = n1 % 2;
int imod2 = n2 % 2;
cout << imod1 << '\n'; // 1
cout << imod2 << '\n'; // 0 

[R0mai]

double dmod1 = (double)(n1 % 2);
double dmod2 = (double)(n2 % 2);

You are mod-ing two integers THEN casting them to double.

[Alan]

well, thats wrong but so is this...

a%10^n

say

a = 1234;
n = 2;

1234%102 ->
1234%100 ->
?????????

the remainder is 12.34, but in c++ this would = 1...

[helios]

Oh, wow.
http://en.wikipedia.org/wiki/Remainder

[Alan]

so the remainder is 34...

[Haptic]

@Alan, it shows 34 here...