maximum value of int

问题

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

回答1:

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;

回答2:

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.

回答3:

#include <climits>
#include <iostream>
using namespace std;

int main() {
  cout << INT_MAX << endl;
}

回答4:

Here is a macro I use to get the maximum value for signed integers, which is independent of the size of the signed integer type used, and for which gcc -Woverflow won't complain

#define SIGNED_MAX(x) (~(-1 << (sizeof(x) * 8 - 1)))

int a = SIGNED_MAX(a);
long b = SIGNED_MAX(b);
char c = SIGNED_MAX(c); /* if char is signed for this target */
short d = SIGNED_MAX(d);
long long e = SIGNED_MAX(e);

回答5:

Why not write a piece of code like:

int  max_neg = ~(1 << 31);
int  all_ones = -1;
int max_pos = all_ones & max_neg;

回答6:

O.K. I neither have rep to comment on previous answer (of Philippe De Muyter) nor raise it's score, hence a new example using his define for SIGNED_MAX trivially extended for unsigned types:

// We can use it to define limits based on actual compiler built-in types also: 
#define INT_MAX   SIGNED_MAX(int)
// based on the above, we can extend it for unsigned types also:
#define UNSIGNED_MAX(x) (  (SIGNED_MAX(x)<<1) | 1 ) // We reuse SIGNED_MAX
#define UINT_MAX  UNSIGNED_MAX(unsigned int) // on ARM: 4294967295
// then we can have:
unsigned int width = UINT_MAX;

Unlike using this or that header, here we use the real type from the compiler.

回答7:

For the specific maximum value of int, I usually write the hexadecimal notation:

int my_max_int = 0x7fffffff;

instead of the irregular decimal value:

int my_max_int = 2147483647;

回答8:

What about (1 << (8*sizeof(int)-2)) - 1 + (1 << (8*sizeof(int)-2)). This is the same as 2^(8*sizeof(int)-2) - 1 + 2^(8*sizeof(int)-2).

If sizeof(int) = 4 => 2^(8*4-2) - 1 + 2^(8*4-2) = 2^30 - 1 + 20^30 = (2^32)/2 - 1 [max signed int of 4 bytes].

You can't use 2*(1 << (8*sizeof(int)-2)) - 1 because it will overflow, but (1 << (8*sizeof(int)-2)) - 1 + (1 << (8*sizeof(int)-2)) works.

回答9:

#include <iostrema>

int main(){
    int32_t maxSigned = -1U >> 1;
    cout << maxSigned << '\n';
    return 0;
}

It might be architecture dependent but it does work at least in my setup.

来源：https://stackoverflow.com/questions/1855459/maximum-value-of-int