I am new to C++.I was going through a C++ book and it says
const int i[] = { 1, 2, 3, 4 };
float f[i[3]]; // Illegal
It says the declaration of the float variable is invalid during compilation.Why is that?
Suppose if we use
int i = 3;
float f[i];
It works.
What is the problem with the first situation?
Thanks.
So the first is illegal because an array must have a compile-time known bound, and i[3], while strictly speaking known at compile time, does not fulfill the criteria the language sets for "compile-time known".
The second is also illegal for the same reason.
Both cases, however, will generally be accepted by GCC because it supports C99-style runtime-sized arrays as an extension in C++. Pass the -pedantic flag to GCC to make it complain.
Edit: The C++ standard term is "integral constant expression", and things qualifying as such are described in detail in section 5.19 of the standard. The exact rules are non-trivial and C++11 has a much wider range of things that qualify due to constexpr, but in C++98, the list of legal things is, roughly:
integer literals
simple expressions involving only constants
non-type template parameters of integral type
variables of integral type declared as const and initialized with a constant expression
Your second example doesn't work and it shouldn't work.
i must be constant. This works
const int i = 3;
float f[i];
Just to expound on Sebastian's answer:
When you create a static array, the compiler must know how much space it needs to reserve. That means the array size must be known at compile-time. In other words, it must be a literal or a constant:
const int SIZE = 3;
int arr[SIZE]; // ok
int arr[3]; // also ok
int size = 3;
int arr[size]; // Not OK
Since the value of size could be different by the time the array is created, the oompiler won't know how much space to reserve for the array. If you declare it as const, it knows the value will not change, and can reserve the proper amount of space.
If you need an array of a variable size, you will need to create it dynamically using new (and make sure to clean it up with delete when you are done with it).
For arrays with lengths known only at runtime in C++ we have std::vector<T>. For builtin arrays the size must be known at compile-time. This is also true for C++11, although the much older C99-standard already supports dynamic stack arrays. See also the accepted answer of Why doesn't C++ support dynamic arrays on the stack?
Related
There is a problem when I want to define a complex array:
#include<complex.h>
int main(){
int matrix=1000;
std::complex<double> y[matrix];
}
The error is "Variable length array of non-POD element type 'std::complex'
Is there something wrong with the definition of array here?
This kind of array only works with a length that is a constant expression, i.e. the length must be known at compile time.
To get a array of variable length, use an std::vector<std::complex<double>> y (matrix);
You should use std::vector (or std::array in some cases) over C-style arrays anyway.
You can't statically allocate a C++ array with size being a regular variable, since the value of matrix is not known until the program is executed. Try dynamically allocating your array:
std::complex<double> y = new std::complex<double>[matrix]
When you are doing using it, call:
delete[] y
The size of arrays must be know at compile time. It must be a constant expression. The value of matrix is only known at runtime. You must make matrix a constant to work.
const int matrix=1000;
The other way around is to use a vector whose size is variable and is initialized at runtime.
int matrix=1000;
std::vector<std::complex<double>> y(matrix);
C++ doesn't allow variable length arrays, either do it dynamically or use a vector.
Your compiler thinks that you are declaring a variable-length array, since matrix is non-const. Just make it constant and things should work:
const int matrix = 1000;
std::complex<double> y[matrix];
The error stems from the fact that variable-length arrays are only allowed for "dumb" data types, e.g. int/char/void* and structs, but not classes like std::complex.
This is what I found during my learning period:
#include<iostream>
using namespace std;
int dis(char a[1])
{
int length = strlen(a);
char c = a[2];
return length;
}
int main()
{
char b[4] = "abc";
int c = dis(b);
cout << c;
return 0;
}
So in the variable int dis(char a[1]) , the [1] seems to do nothing and doesn't work at
all, because I can use a[2]. Just like int a[] or char *a. I know the array name is a pointer and how to convey an array, so my puzzle is not about this part.
What I want to know is why compilers allow this behavior (int a[1]). Or does it have other meanings that I don't know about?
It is a quirk of the syntax for passing arrays to functions.
Actually it is not possible to pass an array in C. If you write syntax that looks like it should pass the array, what actually happens is that a pointer to the first element of the array is passed instead.
Since the pointer does not include any length information, the contents of your [] in the function formal parameter list are actually ignored.
The decision to allow this syntax was made in the 1970s and has caused much confusion ever since...
The length of the first dimension is ignored, but the length of additional dimensions are necessary to allow the compiler to compute offsets correctly. In the following example, the foo function is passed a pointer to a two-dimensional array.
#include <stdio.h>
void foo(int args[10][20])
{
printf("%zd\n", sizeof(args[0]));
}
int main(int argc, char **argv)
{
int a[2][20];
foo(a);
return 0;
}
The size of the first dimension [10] is ignored; the compiler will not prevent you from indexing off the end (notice that the formal wants 10 elements, but the actual provides only 2). However, the size of the second dimension [20] is used to determine the stride of each row, and here, the formal must match the actual. Again, the compiler will not prevent you from indexing off the end of the second dimension either.
The byte offset from the base of the array to an element args[row][col] is determined by:
sizeof(int)*(col + 20*row)
Note that if col >= 20, then you will actually index into a subsequent row (or off the end of the entire array).
sizeof(args[0]), returns 80 on my machine where sizeof(int) == 4. However, if I attempt to take sizeof(args), I get the following compiler warning:
foo.c:5:27: warning: sizeof on array function parameter will return size of 'int (*)[20]' instead of 'int [10][20]' [-Wsizeof-array-argument]
printf("%zd\n", sizeof(args));
^
foo.c:3:14: note: declared here
void foo(int args[10][20])
^
1 warning generated.
Here, the compiler is warning that it is only going to give the size of the pointer into which the array has decayed instead of the size of the array itself.
The problem and how to overcome it in C++
The problem has been explained extensively by pat and Matt. The compiler is basically ignoring the first dimension of the array's size effectively ignoring the size of the passed argument.
In C++, on the other hand, you can easily overcome this limitation in two ways:
using references
using std::array (since C++11)
References
If your function is only trying to read or modify an existing array (not copying it) you can easily use references.
For example, let's assume you want to have a function that resets an array of ten ints setting every element to 0. You can easily do that by using the following function signature:
void reset(int (&array)[10]) { ... }
Not only this will work just fine, but it will also enforce the dimension of the array.
You can also make use of templates to make the above code generic:
template<class Type, std::size_t N>
void reset(Type (&array)[N]) { ... }
And finally you can take advantage of const correctness. Let's consider a function that prints an array of 10 elements:
void show(const int (&array)[10]) { ... }
By applying the const qualifier we are preventing possible modifications.
The standard library class for arrays
If you consider the above syntax both ugly and unnecessary, as I do, we can throw it in the can and use std::array instead (since C++11).
Here's the refactored code:
void reset(std::array<int, 10>& array) { ... }
void show(std::array<int, 10> const& array) { ... }
Isn't it wonderful? Not to mention that the generic code trick I've taught you earlier, still works:
template<class Type, std::size_t N>
void reset(std::array<Type, N>& array) { ... }
template<class Type, std::size_t N>
void show(const std::array<Type, N>& array) { ... }
Not only that, but you get copy and move semantic for free. :)
void copy(std::array<Type, N> array) {
// a copy of the original passed array
// is made and can be dealt with indipendently
// from the original
}
So, what are you waiting for? Go use std::array.
It's a fun feature of C that allows you to effectively shoot yourself in the foot if you're so inclined. I think the reason is that C is just a step above assembly language. Size checking and similar safety features have been removed to allow for peak performance, which isn't a bad thing if the programmer is being very diligent. Also, assigning a size to the function argument has the advantage that when the function is used by another programmer, there's a chance they'll notice a size restriction. Just using a pointer doesn't convey that information to the next programmer.
First, C never checks array bounds. Doesn't matter if they are local, global, static, parameters, whatever. Checking array bounds means more processing, and C is supposed to be very efficient, so array bounds checking is done by the programmer when needed.
Second, there is a trick that makes it possible to pass-by-value an array to a function. It is also possible to return-by-value an array from a function. You just need to create a new data type using struct. For example:
typedef struct {
int a[10];
} myarray_t;
myarray_t my_function(myarray_t foo) {
myarray_t bar;
...
return bar;
}
You have to access the elements like this: foo.a[1]. The extra ".a" might look weird, but this trick adds great functionality to the C language.
To tell the compiler that myArray points to an array of at least 10 ints:
void bar(int myArray[static 10])
A good compiler should give you a warning if you access myArray [10]. Without the "static" keyword, the 10 would mean nothing at all.
This is a well-known "feature" of C, passed over to C++ because C++ is supposed to correctly compile C code.
Problem arises from several aspects:
An array name is supposed to be completely equivalent to a pointer.
C is supposed to be fast, originally developerd to be a kind of "high-level Assembler" (especially designed to write the first "portable Operating System": Unix), so it is not supposed to insert "hidden" code; runtime range checking is thus "forbidden".
Machine code generrated to access a static array or a dynamic one (either in the stack or allocated) is actually different.
Since the called function cannot know the "kind" of array passed as argument everything is supposed to be a pointer and treated as such.
You could say arrays are not really supported in C (this is not really true, as I was saying before, but it is a good approximation); an array is really treated as a pointer to a block of data and accessed using pointer arithmetic.
Since C does NOT have any form of RTTI You have to declare the size of the array element in the function prototype (to support pointer arithmetic). This is even "more true" for multidimensional arrays.
Anyway all above is not really true anymore :p
Most modern C/C++ compilers do support bounds checking, but standards require it to be off by default (for backward compatibility). Reasonably recent versions of gcc, for example, do compile-time range checking with "-O3 -Wall -Wextra" and full run-time bounds checking with "-fbounds-checking".
C will not only transform a parameter of type int[5] into *int; given the declaration typedef int intArray5[5];, it will transform a parameter of type intArray5 to *int as well. There are some situations where this behavior, although odd, is useful (especially with things like the va_list defined in stdargs.h, which some implementations define as an array). It would be illogical to allow as a parameter a type defined as int[5] (ignoring the dimension) but not allow int[5] to be specified directly.
I find C's handling of parameters of array type to be absurd, but it's a consequence of efforts to take an ad-hoc language, large parts of which weren't particularly well-defined or thought-out, and try to come up with behavioral specifications that are consistent with what existing implementations did for existing programs. Many of the quirks of C make sense when viewed in that light, particularly if one considers that when many of them were invented, large parts of the language we know today didn't exist yet. From what I understand, in the predecessor to C, called BCPL, compilers didn't really keep track of variable types very well. A declaration int arr[5]; was equivalent to int anonymousAllocation[5],*arr = anonymousAllocation;; once the allocation was set aside. the compiler neither knew nor cared whether arr was a pointer or an array. When accessed as either arr[x] or *arr, it would be regarded as a pointer regardless of how it was declared.
One thing that hasn't been answered yet is the actual question.
The answers already given explain that arrays cannot be passed by value to a function in either C or C++. They also explain that a parameter declared as int[] is treated as if it had type int *, and that a variable of type int[] can be passed to such a function.
But they don't explain why it has never been made an error to explicitly provide an array length.
void f(int *); // makes perfect sense
void f(int []); // sort of makes sense
void f(int [10]); // makes no sense
Why isn't the last of these an error?
A reason for that is that it causes problems with typedefs.
typedef int myarray[10];
void f(myarray array);
If it were an error to specify the array length in function parameters, you would not be able to use the myarray name in the function parameter. And since some implementations use array types for standard library types such as va_list, and all implementations are required to make jmp_buf an array type, it would be very problematic if there were no standard way of declaring function parameters using those names: without that ability, there could not be a portable implementation of functions such as vprintf.
It's allowed for compilers to be able to check whether the size of array passed is the same as what expected. Compilers may warn an issue if it's not the case.
double rainPerMonth(const int YEARS)
{
int monthYear[MONTHS][YEARS];
// ...
}
Visual Studio shows a squiggly line underneath the array declaration, saying that YEARS must be a constant when I'm creating the array. Is this an IDE issue because the variable has yet to be initialized, or am I writing this incorrectly?
MONTHS is already declared globally.
An array size must be a constant expression - that is, a value known at compile time. (Some compilers offer C-style variable-length arrays as a non-standard extension, but I don't think Visual C++ does. Even if it does, it's better not to rely on such extensions.)
A function argument isn't known at compile time, so can't be used as an array size. Your best option is here is probably
std::vector<std::array<int, MONTHS>> monthYear(YEARS);
In C++, an array must be sized at compile time. What you are attempting to do is declare one that is sized at runtime. In the function you've declared, YEARS is only constant within the scope of the function. You could call it rainPerMonth(someInt); where someInt is the result of some user input (which shows you that the result is not a compile-time constant).
Variable Length Arrays are an extension to C, but not C++. To do what you want, you can use dynamic memory, or a std::vector.
I think your problem lies in the fact that C++ wants a constant in the sense of compile-time constant to create your variable monthYear. If you pass it as a function, it need not be known at compile time? For example:
const int x=2;
const int y=3;
char xyChoice;
std::cin >> xyChoice;
if (xyChoice == 'x')
rainPerMonth(x);
else
rainPerMonth(y);
I'm unsure, but it seems to me like this would give you a constant int being passed to your function, but the compiler wouldn't know what size to create an array for before runtime?
I'm trying to do something like this:
const int array_size = 5;
string stuff[array_size];
My compiler won't let me compile this, even though array_size is a constant. Is there a way to do this without dealing with dynamic arrays?
Edit: "error C2057: expected constant expression"
I have answered this question assuming you are either coding in C or C++. If you are using a different language, this answer doesn't apply. However, you should update your question with the language you are trying to use.
Consider the following program:
int main () {
const int size = 5;
int x[size];
return 0;
}
This will compile in both C++ and C.99, but not C.89. In C.99, variable length arrays were introduced, and so locally scoped arrays can take on a size specified by a variable. However, arrays at file scope in C.99 cannot take a variable size parameter, and in C.89, all array definitions have to have a non variable size.
If you are using C.89, or defining a file scope array in C.99, you can use an enum to name your constant value. The enum can then be used to size the array definition. This is not necessary for C++ however, which allows a const integer type initialized by a literal to be used to size an array declaration.
enum { size = 5 };
int x[size];
int main () { return 0; }
#define array_size 5
string stuff[array_size];
You can use e.g. a vector or the new keyword to allocate memory dynamically, because declared arrays can not have runtime sizes.
Only thing I can think of is that you defined another array_size variable in your code, which is not a compile time constant and hides the original array_size.
array_size is not treated as a compile time constant. Constness added just makes sure that programmer can not modify it. If tried to modify accidentally, compiler will bring to your attention.
Size of an array needs to be a compile constant. Seems like your compiler is not supporting Variable Length Array. You can #define the size of the array instead which is treated as a constant expression.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Is there a way to initialize an array with non-constant variables? (C++)
I have the following code:
vector<vector<vec2>> vinciP;
int myLines = -1;
myLines = drawPolyLineFile("vinci.dat", vinciP);
if (myLines > -1)
{
cout << "\n\nSUCCESS";
vec2 vPoints[myLines];
for (int i = 0; i < NumPoints; ++i)
{
vPoints[i] = vinciP[0][i];
}
}
I'm getting an error on the line 'vec2 vPoints[myLines];' that says expressions must have a constant value. I don't understand why I'm getting this error, any help?
Is it because myLines could be negative? idk.
vec2 vPoints[myLines];
Since myLines is not a const expression ((which means, it is not known at compile-time), so the above code declares a variable length array which is not allowed in C++. Only C99 has this feature. Your compiler might have this as an extension (but that is not Standard C++).
The solution to such commom problem is : use std::vector<T> as:
std::vector<vec2> vPoints(myLines);
It should work now.
Is it because myLines could be negative?
No, It is because myLines is not a compile time constant.
Explanation:
vec2 vPoints[myLines];
Creates an array of variable length, where myLines value will be determined at Run-time.
Variable length arrays are not allowed in C++. It was a feature introduced in C99, and C++ Standard does not support it. Some C++ compilers support it as an extension though but it is nevertheless non standard conforming.
For C++ size of an array should be known at compile time and hence must be compile time constant. myLines is not a compile time constant and hence the error.
You should use a std::vector
vec2 vPoints[myLines];
Array size must be a compile time constant. myLines is not a compile time constant. Instead, allocate the memory using new or even better to use std::vector.
C++ does not have variable-length arrays. The size of an array must be determined at compile-time. The value of myLines is only known at runtime, so this won't work.
To have arrays whose size is only known at runtime, use std::vector.
std::vector<vec2> vPoints(myLines);
You are getting that error because static arrays need a static (constant) size. Since the number of components in your vPoints is dynamic, consider using a dynamic array instead. Or better yet stick with vector.