Declaring my int array in C

Question

I want to initialize an int array with constant values, with all the 2^n values from 2^0 to 2^31 and I want to know which method below is right and if it gives the result I want and if there are easier or shorter methods to initialize it.

static const char     two_n[32];

two_n[32] = {1, 2 ,4 ,8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096,
               8192, 16384, 32768, 65535, 131070, 262140, 524280, 1048560,
               2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728,
               268435456, 536870912, 1073741824, 2147483648};

or

static const char     *two_n[32];

two_n[32] = {1, 2 ,4 ,8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096,
               8192, 16384, 32768, 65535, 131070, 262140, 524280, 1048560,
               2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728,
               268435456, 536870912, 1073741824, 2147483648};

or

static const int     two_n[32];

two_n[32] = {1, 2 ,4 ,8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096,
               8192, 16384, 32768, 65535, 131070, 262140, 524280, 1048560,
               2097152, 4194304, 8388608, 16777216, 33554432, 67108864, 134217728,
               268435456, 536870912, 1073741824, 2147483648};

None of these will work. Did you just try the code and see the results? You need to review your understanding of arrays. Try something simpler. — john elemans
– john elemans, Commented Nov 27, 2016 at 21:49
65535 is wrong along with several more after that. You'd be better off filling the array with a simple for loop when the program starts. Or you could use hexadecimal notation for the constants, since the hexadecimal values follow a nice obvious pattern. — user3386109
– user3386109, Commented Nov 27, 2016 at 21:52
Thanks for the comments, I'm in an it school where using the for loops is forbidden in first year, and I can't use hexadecimal because I need to do a program that sorts numbers using Radix, and I am converting numbers to binary base then sorting using ascending binary order. — user7217454
– user7217454, Commented Nov 27, 2016 at 22:26
the values for the powers of 2 are given by: static const int two_n[32] = { 1<<0, 1<<2, 1<<3, 1<<4,..... 1<<31 };` — user3629249
– user3629249, Commented Nov 29, 2016 at 12:07
Note: after a const array is declared, it cannot be assigned to. (otherwise it would not be const.) — user3629249
– user3629249, Commented Nov 29, 2016 at 12:09

chqrlie · Accepted Answer · 2016-11-27 21:59:48Z

3

Neither of the above because both the types of the arrays and the initializers are incorrect. Use uint32_t or unsigned long as these are guaranteed to be at least 32-bit wide and able to hold the value of 2³¹, which might be beyond the range of int:

static const unsigned long two_n[32] = {
    1UL <<  0, 1UL <<  1, 1UL <<  2, 1UL <<  3,
    1UL <<  4, 1UL <<  5, 1UL <<  6, 1UL <<  7,
    1UL <<  8, 1UL <<  9, 1UL << 10, 1UL << 11,
    1UL << 12, 1UL << 13, 1UL << 14, 1UL << 15,
    1UL << 16, 1UL << 17, 1UL << 18, 1UL << 19,
    1UL << 20, 1UL << 21, 1UL << 22, 1UL << 23,
    1UL << 24, 1UL << 25, 1UL << 26, 1UL << 27,
    1UL << 28, 1UL << 29, 1UL << 30, 1UL << 31
};

answered Nov 27, 2016 at 21:59

chqrlie

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2 Comments

user7217454 Over a year ago

Thanks for the comment, I'll use long type and not unsigned because I'll also need negative values.

chqrlie Over a year ago

@MohammedBRAHMI: If long type is 32-bit wide, it cannot hold 2^31. You would need long long which is guaranteed to hold at least up to 2^63-1

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