const variable not recognized as array dimension

Question

long AnsiString::pos(const char* plainString) const {
const size_t patternLength = strlen(plainString);
if (patternLength == 0) return -1;

size_t stringLength = count;
int partialMatch[patternLength]; // int* partialMatch = new int[patternLength];

KMPBuildPartialMatchTable(plainString, partialMatch);

int currentStringCharacter = 0;
int currentPatternCharacter = 0;
while (currentStringCharacter < stringLength) {
    if (currentPatternCharacter == -1) {
        currentStringCharacter++;
        currentPatternCharacter = 0;
    }
    else if (items[currentStringCharacter] == plainString[currentPatternCharacter]) {
        currentStringCharacter++;
        currentPatternCharacter++;
        if (currentPatternCharacter == patternLength) return currentStringCharacter - currentPatternCharacter;
    } else {
        currentPatternCharacter = partialMatch[currentPatternCharacter];
    }
}

// delete(partialMatch);

return -1;
 }

I get an error in the implementaation of this claas method using visual c++.

int partialMatch[ patternLength ] ; // expression must have a constant value

(I'm using VS in other language so you could find some differences).

I declared patternLength as a constant as you can see. A solution is commented in the ccode but i don't want to use dynamic memory allocation. Some idea ?

Array size must a *constant-expression*. For short a *constant-expression* is an expression evaluable at compile-time. — Oliv, Sep 08 '18 at 17:14
Removed the "duplicates" : they didn't address the question of why a `const` variable may not suffice in C++. (OP understands that non-const variables cannot be array dimensions, which is what those links covered). — M.M, May 14 '19 at 02:07

R Sahu · Accepted Answer · 2018-09-08T17:28:16.953

Array sizes must be known at compile time.

A const variable does not ensure that. The const qualifier ensures that the variable cannot be modified once it is initialized.

It is possible for the value of const variable to be known at compile time. If a compiler can detect that, then the variable can be used to define the size of an array.

More often, the value of a const variable is not known at compile time. It is initialized with a value at run time, which can't be changed after the variable is initialized. That does not make it fit for use to define the size of an array.

If you want to be able to use the variable at compile time, use constexpr instead. The compiler will do its best to evaluate the value at compile time. It will fail if the value of the variable cannot be evaluated at compile time.

Minor spelling fix: `constexpr` – Derek T. Jones Sep 08 '18 at 17:27 — Derek T. Jones, Sep 08 '18 at 17:27
@DerekT.Jones, sharp eyes. It's fixed. – R Sahu Sep 08 '18 at 17:28 — R Sahu, Sep 08 '18 at 17:28

Swordfish · Answer 2 · 2018-09-08T17:34:52.577

2

The size N in an array declaration T[N] must be a compile-time constant-expression.

std::size_t const a{42}; // is one,

std::size_t foo{std::rand() % 100};
std::size_t const b{foo}; // is not, because it depends on foo

edited Sep 08 '18 at 17:34

answered Sep 08 '18 at 17:14

Swordfish

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Actually, all of these will work to define array dimensions because the compiler can evaluate them as constexpr at compile time. – Derek T. Jones Sep 08 '18 at 17:26
@DerekT.Jones no longer. – Swordfish Sep 08 '18 at 17:35
Yes, it’s much better example now! Thanks. – Derek T. Jones Sep 09 '18 at 16:38

Ron · Answer 3 · 2018-09-08T17:34:00.897

Marking something const does not make it a constant expression per se. It makes it a read only. The right hand side of your statement should satisfy constexpr function requirements, which the strlen(plainString) expression does not. You could make your own function that is evaluated during compile time:

constexpr size_t constexprlength(const char* s) {
    return strlen(s);
}

and use that instead:

constexpr size_t patternLength = constexprlength(plainString);
int partialMatch[patternLength];

VLAs and character arrays are evil. Use std::vector and std::string instead.

const variable not recognized as array dimension

3 Answers3