maximum value of int


Question

Is there any code to find the maximum value of integer (accordingly to the compiler) in C/C++ like Integer.MaxValue function in java?

1
161
11/26/2015 3:13:40 PM

Accepted Answer

In C++:

#include <limits>

then use

int imin = std::numeric_limits<int>::min(); // minimum value
int imax = std::numeric_limits<int>::max();

std::numeric_limits is a template type which can be instantiated with other types:

float fmin = std::numeric_limits<float>::min(); // minimum positive value
float fmax = std::numeric_limits<float>::max();

In C:

#include <limits.h>

then use

int imin = INT_MIN; // minimum value
int imax = INT_MAX;

or

#include <float.h>

float fmin = FLT_MIN;  // minimum positive value
double dmin = DBL_MIN; // minimum positive value

float fmax = FLT_MAX;
double dmax = DBL_MAX;
298
1/17/2012 9:26:53 AM

I know it's an old question but maybe someone can use this solution:

int size = 0; // Fill all bits with zero (0)
size = ~size; // Negate all bits, thus all bits are set to one (1)

So far we have -1 as result 'till size is a signed int.

size = (unsigned int)size >> 1; // Shift the bits of size one position to the right.

As Standard says, bits that are shifted in are 1 if variable is signed and negative and 0 if variable would be unsigned or signed and positive.

As size is signed and negative we would shift in sign bit which is 1, which is not helping much, so we cast to unsigned int, forcing to shift in 0 instead, setting the sign bit to 0 while letting all other bits remain 1.

cout << size << endl; // Prints out size which is now set to maximum positive value.

We could also use a mask and xor but then we had to know the exact bitsize of the variable. With shifting in bits front, we don't have to know at any time how many bits the int has on machine or compiler nor need we include extra libraries.


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