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I have a question about template function. This is my first time using template. I may lack of some basis. I am confusing about the type of variables I should use. What is the meaning of int N after the template? I regard N as a flexible integer. Is A[N][N] 2d array variable or a pointer? What about b[N]? I have an example using all inputs as non-pointer ones. But the hint from the Xcode tells me all variable are pointer. Here is the hint "LUsolve(double (*A)[N], const double *b, double *x)". But I tried my pointers variable, not working. The function LUfactorize and LUsolve_internal are also template functions. It is too long. I don't post all code.

template < int N > bool LUsolve( double A[N][N], const double b[N], double x[N] ) { double B[N][N]; int i, j, p[N], status; for (i = 0; i < N; i++) { for (j = 0; j < N; j++) { B[i][j] = A[i][j]; } } status = LUfactorize(B, p); if (status != 0) { printf("Failed in the LU factorization.\n"); return false; } LUsolve_internal(B, p, b, x); return true; } template <int N> int LUfactorize(double A[N][N], int p[N]) template <int N> void LUsolve_internal(double A[N][N], const int p[N], const double b[N], double x[N])

Here is the example I work on. The error is "No matching function for call to 'LUsolve'". How can I fix it?

bool cubicspline (double const *knots , double const *knots_value, double *coef, int const N) { double x_i = 0; double x_Km1 = 0; double x_K = 0; double d_i = 0; double d_Km1 = 0; double A [N][N]; for ( int i = 0; i < N; i++) { A[i][0] = 1; A[i][1] = knots[i]; for ( int j = 2; j < N; j++) { x_i = ( (knots[i] > knots[j-2]) ? (knots[i] - knots[j-2]) : 0 ); x_Km1 = ( (knots[i] > knots [N-2]) ? (knots[i] - knots[N-2]) : 0 ); x_K = ( (knots[i] > knots [N-1]) ? (knots[i] - knots[N-1]) : 0 ); d_i = (x_i * x_i * x_i - x_K * x_K * x_K) / ( knots[N-1] - knots[j-2] ); d_Km1 = ( x_Km1 * x_Km1 * x_Km1 - x_K * x_K * x_K) / ( knots[N-1] - knots[N-2] ); A[i][j] = d_i - d_Km1; } } bool status = LUsolve( A , knots_value, coef); // Here is the error.
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    Your variable names are quite confusing, specially using both lower and upper case of the same later (see b and B). The time spent writing a few extra characters will repay itself countless times by the time you and everyone else will save trying to figure out what your code is doing. Commented Jul 17, 2017 at 15:12
  • The language doesn't allow passing arrays in function parameters, so they all decay to pointers. Commented Jul 17, 2017 at 15:13
  • B is a matrix and b is the right hand side of the linear system, e.g. Bu=b. Commented Jul 17, 2017 at 15:14
  • Pretty simple: N will be replaced with the integer constant you provide at instantiation. Commented Jul 17, 2017 at 15:14
  • Why the array could be used in the example? Commented Jul 17, 2017 at 15:20

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Having template< int N > in a sense means that N is another parameter to the function - it's just like a normal variable. The big difference is that it must be specified at compile time. This lets you declare arrays of fixed size N when you can't normally declare an array of size determined by a variable - the only reason it works here is because N is known at compile-time, so it's "like" writing int x[5].

See here for a similar question, and the linked wikipedia article on Template Metaprogramming for a better appreciation of the compile-time nature.

If as you said, this is your first time using templates, you should probably start with more "normal" usecases (as generic parameters) to get a feel for how they're compile-time, before using them like this.

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About template function

It is a function template. You get a template function by instantiating the function template.

What is the meaning of int N after the template?

It is a (non-type) template parameter. When the template is instantiated, the values for the template parameters (and types for template type parameters, which your template does not have) must be provided - either explicitly, or deduced from the non-template arguments by the compiler)

Is A[N][N] 2d array variable or a pointer?

It is a pointer. But only because it is a function argument. If it was a declaration of an array that was not a function parameter, then it would have been an array variable.

You cannot pass arrays by value into functions, and an array argument declaration is sneakily (and confusingly in my opinion) replaced with a pointer. The type is actually double (*A)[N] i.e. a pointer to an array of N doubles.

Here is the example I said. The inputs are not pointers.

Like above, your inputs really are pointers. Function parameter const double p[2] actually means double const *p.

Why the array could be used in the example?

Because arrays implicitly decay to pointer to first element.

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13 Comments

Thank you for the answer. I have changed the example to mine. Can you give some advise?
@Yue did the compiler not mention any candidates for the function? I suspect that you forgot to define the template before calling it.
@Yue You can however use references to arrays, e.g. double (&A)[N][N]
@user2079303 The complier gives the candidate function as 'LUsolve'. The first function is in the LU.h file. My example is in another header file. I have included LU.h. What do you mean by define template? Define template in my example function?
@Yue does it not mention anything about not being able to convert arguments? Read the whole error.
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