This is not specific to any programming language, the problem is "find the index of a specified value in an array of n numbers.
Now my question is, in the code below can you declare an array how I have done it.
{int n;
read(n);
int array[n];
......
or is this allowed?
{int n; array[n];
read(n)
I'm thinking the first one is correct.
Thanks in advance.
Converted from a comment as suggested by Merlyn Morgan-Graham
The way an array is declared depends on what language you use. If you are writing pseudo-code you can decide it yourself as long as it communicates the intent and the desired result.
The array can be declared as array = [], int[] array = new int[], int array[], array = array(), ´array = {}` etc. In some languages you have to declare the size of the array beforehand and in some languages the arrays expand when needed
In terms of syntax - that would certainly be programming language dependent. But assuming the programming language behaves more or less statically and treats arrays as statically allocated blocks in memory (rather than vectors, etc.), etc. then the first option must be correct as only after n is read a static array can be allocated.
Of course the first one is correct. In the second one when you declare the array, n is not yet set. So it is not correct.
Normally when creating an array you need to know the size before-hand. Whether you know the value at compile-time or run-time can be dependent on your language/project requirements, but it must be known before you can decide to create an array of that size. (i.e. the first solution is correct)
Related
Because in C the array length has to be stated when the array is defined, would it be acceptable practice to use the first element as the length, e.g.
int arr[9]={9,0,1,2,3,4,5,6,7};
Then use a function such as this to process the array:
int printarr(int *ARR) {
for (int i=1; i<ARR[0]; i++) {
printf("%d ", ARR[i]);
}
}
I can see no problem with this but would prefer to check with experienced C programmers first. I would be the only one using the code.
Well, it's bad in the sense that you have an array where the elements does not mean the same thing. Storing metadata with the data is not a good thing. Just to extrapolate your idea a little bit. We could use the first element to denote the element size and then the second for the length. Try writing a function utilizing both ;)
It's also worth noting that with this method, you will have problems if the array is bigger than the maximum value an element can hold, which for char arrays is a very significant limitation. Sure, you can solve it by using the two first elements. And you can also use casts if you have floating point arrays. But I can guarantee you that you will run into hard traced bugs due to this. Among other things, endianness could cause a lot of issues.
And it would certainly confuse virtually every seasoned C programmer. This is not really a logical argument against the idea as such, but rather a pragmatic one. Even if this was a good idea (which it is not) you would have to have a long conversation with EVERY programmer who will have anything to do with your code.
A reasonable way of achieving the same thing is using a struct.
struct container {
int *arr;
size_t size;
};
int arr[10];
struct container c = { .arr = arr, .size = sizeof arr/sizeof *arr };
But in any situation where I would use something like above, I would probably NOT use arrays. I would use dynamic allocation instead:
const size_t size = 10;
int *arr = malloc(sizeof *arr * size);
if(!arr) { /* Error handling */ }
struct container c = { .arr = arr, .size = size };
However, do be aware that if you init it this way with a pointer instead of an array, you're in for "interesting" results.
You can also use flexible arrays, as Andreas wrote in his answer
In C you can use flexible array members. That is you can write
struct intarray {
size_t count;
int data[]; // flexible array member needs to be last
};
You allocate with
size_t count = 100;
struct intarray *arr = malloc( sizeof(struct intarray) + sizeof(int)*count );
arr->count = count;
That can be done for all types of data.
It makes the use of C-arrays a bit safer (not as safe as the C++ containers, but safer than plain C arrays).
Unforntunately, C++ does not support this idiom in the standard.
Many C++ compilers provide it as extension though, but it is not guarantueed.
On the other hand this C FLA idiom may be more explicit and perhaps more efficient than C++ containers as it does not use an extra indirection and/or need two allocations (think of new vector<int>).
If you stick to C, I think this is a very explicit and readable way of handling variable length arrays with an integrated size.
The only drawback is that the C++ guys do not like it and prefer C++ containers.
It is not bad (I mean it will not invoke undefined behavior or cause other portability issues) when the elements of array are integers, but instead of writing magic number 9 directly you should have it calculate the length of array to avoid typo.
#include <stdio.h>
int main(void) {
int arr[9]={sizeof(arr)/sizeof(*arr),0,1,2,3,4,5,6,7};
for (int i=1; i<arr[0]; i++) {
printf("%d ", arr[i]);
}
return 0;
}
Only a few datatypes are suitable for that kind of hack. Therefore, I would advise against it, as this will lead to inconsistent implementation styles across different types of arrays.
A similar approach is used very often with character buffers where in the beginning of the buffer there is stored its actual length.
Dynamic memory allocation in C also uses this approach that is the allocated memory is prefixed with an integer that keeps the size of the allocated memory.
However in general with arrays this approach is not suitable. For example a character array can be much larger than the maximum positive value (127) that can be stored in an object of the type char. Moreover it is difficult to pass a sub-array of such an array to a function. Most of functions that designed to deal with arrays will not work in such a case.
A general approach to declare a function that deals with an array is to declare two parameters. The first one has a pointer type that specifies the initial element of an array or sub-array and the second one specifies the number of elements in the array or sub-array.
Also C allows to declare functions that accepts variable length arrays when their sizes can be specified at run-time.
It is suitable in rather limited circumstances. There are better solutions to the problem it solves.
One problem with it is that if it is not universally applied, then you would have a mix of arrays that used the convention and those that didn't - you have no way of telling if an array uses the convention or not. For arrays used to carry strings for example you have to continually pass &arr[1] in calls to the standard string library, or define a new string library that uses "Pascal strings" rather then "ASCIZ string" conventions (such a library would be more efficient as it happens),
In the case of a true array rather then simply a pointer to memory, sizeof(arr) / sizeof(*arr) will yield the number of elements without having to store it in the array in any case.
It only really works for integer type arrays and for char arrays would limit the length to rather short. It is not practical for arrays of other object types or data structures.
A better solution would be to use a structure:
typedef struct
{
size_t length ;
int* data ;
} intarray_t ;
Then:
int data[9] ;
intarray_t array{ sizeof(data) / sizeof(*data), data } ;
Now you have an array object that can be passed to functions and retain the size information and the data member can be accesses directly for use in third-party or standard library interfaces that do not accept the intarray_t. Moreover the type of the data member can be anything.
Obviously NO is the answer.
All programming languages has predefined functions stored along with the variable type. Why not use them??
In your case is more suitable to access count /length method instead of testing the first value.
An if clause sometimes take more time than a predefined function.
On the first look seems ok to store the counter but imagine you will have to update the array. You will have to do 2 operations, one to insert other to update the counter. So 2 operations means 2 variables to be changed.
For statically arrays might be ok to have them counter then the list, but for dinamic ones NO NO NO.
On the other hand please read programming basic concepts and you will find your idea as a bad one, not complying with programming principles.
include <stdio.h>
int main() {
int num = 10;
int arr[num];
for(int i = 0; i < num; i++){
arr[num] = i+1;
}
}
Some colleague of mine says that this code is not correct and that it is illegal. However, when I am running it, it is working without any errors. And he does not know how to explain why it is working and why I should not code like this. Can you please help me. I am a beginner and I want to learn C.
If you want to dynamically allocate an array of length n ints, you'll need to use either malloc or calloc. Calloc is preferred for array allocation because it has a built in multiplication overflow check.
int num = 10;
int *arr = calloc(num, sizeof(*arr));
//Do whatever you need to do with arr
free(arr);
arr = NULL;
Whenever you allocate memory with malloc or calloc, always remember to free it afterwards, then set the pointer to NULL in order to prevent any accidental, future references, as well as to prevent a double free.
While not necessarily illegal, this code won't do what you intend. When you declare an array, you declare the number of items you want to store, in this instance num. So when you declare num = 10 and arr[num] you get an array that can hold 10 integers. C arrays are indexed from 0, so the indices are 0-9, not 1-10. This is probably what they mean by illegal. Since you are writing to arr[num] or arr[10], you are attempting to use memory beyond the memory allocated for the array.
Additionally, if I understand the intent of the program correctly, you want to fill in the array with the numbers 1-10. To do this, you'd need to access each index individually. You're almost there, the only problem being arr[num] = i + 1;. As mentioned before, it is beyond the end of the array. However, you should probably be using i as your index, so arr[i], because this will access each index, 0-9.
Are you learning C or C++?
Your colleague meant that in that code of yours you are doing something different from what you wanted. It's working because of some additional factors. Because C/C++ standards are evolving and so do compilers as well. Let me show you.
Static array
When you a beginner, it's generally advised to stick to the concept that "a typed array of the compilation-given size" is int arr[N], where N is a constant. You allocate it on the stack and you don't manage it's memory.
In C++11 you can use a constexpr (constant expression), but is still not an arbitrary variable.
In C++14 you can use a "simple expression" for size, but you shouldn't try a lot of it before getting the array concept beforehand. Also, GCC compiler provides an extension to support variable sized arrays, it could be an explanation of "why the code is working at all".
Notice: variable sized arrays are not the same as dynamic arrays. They are not that static arrays from the first chapter of a C/C++ guide book as well.
There also exists a modern approach – std::array<int, 10> but once again, don't start with it.
Dynamic array
When you need to create an array in runtime everything changes. First of all, you allocate it on the heap and either you mange it's memory yourself (if you do not, you get a memory leak, a Pure C way) or use special C++ classes like std::vector. Once again, vectors should be used after getting to know Pure C arrays.
Your colleague must have been meaning something like that:
int* arr = new int[some_variable]; // this is dynamic array allocation
delete[] arr; // in modern C/C++ you can write "delete arr;" as well
So, your compiler made it work in this exact case, but you definitely should not rely on the approach you've tried. It's not an array allocation at all.
TL;DR:
In C++ variable length arrays are not legal
g++ compiler allows variable length arrays, because C99 allows them
Remember that C and C++ are two different languages
The piece of code from the question seems to be not doing what you'd wanted it to do
As others mentioned, it should be arr[i] = i + 1 instead, you are assigning to the same array item all the time otherwise
before you mark this as a duplicate please notice that I'm looking for a more general solution for arrays of arbitrary dimensions. I have read many posts here or in forums about making 2D or 3D arrays of integers but these are specific solutions for specific dimensions. I want a general solution for an array of any dimension.
First I need to have a type of intlist as defined below:
typedef struct{
int l // length of the list
int * e // pointer to the first element of the array
}intlist;
this actually fills the gap in C for treating arrays just as pointers. using this type I can pass arrays to functions without worrying about loosing the size.
then in the next step I want to have a mdintlist as multidimensional dynamically allocated arrays. the type definition should be something like this:
typedef struct Mdintlist{
intlist d // dimension of the array
/* second part */
}mdintlist;
there are several options for the second part. on option is that to have a pointer towards a mdintlist of lower dimension like
struct Mdintlist * c;
the other options is to use void pointers:
void * c;
I don't know how to continue it from here.
P.S. one solution could be to allocate just one block of memory and then call the elements using a function. However I would like to call the elements in array form. something like tmpmdintlist.c[1][2][3]...
Hope I have explained clearly what I want.
P.S. This is an ancient post, but for those who may end up here some of my efforts can be seen in the Cplus repo.
You can't! you can only use the function option in c, because there is no way to alter the language semantics. In c++ however you can overload the [] operator, and even though I would never do such an ugly thing (x[1][2][3] is alread y ugly, if you continue adding "dimensions" it gets really ugly), I think it would be possible.
Well, if you separate the pointers and the array lengths, you end up with much less code.
int *one_dem_array;
size_t one_dem_count[1];
int **two_dem_array;
size_t two_dem_count[2];
int ***three_dem_array;
size_t three_dem_count[3];
This way you can still use your preferred notation.
int num_at_pos = three_dem_array[4][2][3];
given the following function signature:
void readFileData(FILE* fp, double inputMatrix[][], int parameters[])
this doesn't compile.
and the corrected one:
void readFileData(FILE* fp, double inputMatrix[][NUM], int parameters[])
my question is, why does the compiler demands that number of columns will be defined when handling a 2D array in C? Is there a way to pass a 2D array to a function with an unknown dimensions?
thank you
Built-in multi-deminsional arrays in C (and in C++) are implemented using the "index-translation" approach. That means that 2D (3D, 4D etc.) array is laid out in memory as an ordinary 1D array of sufficient size, and the access to the elements of such array is implemented through recalculating the multi-dimensional indices onto a corresponding 1D index. For example, if you define a 2D array of size M x N
double inputMatrix[M][N]
in reality, under the hood the compiler creates an array of size M * N
double inputMatrix_[M * N];
Every time you access the element of your array
inputMatrix[i][j]
the compiler translates it into
inputMatrix_[i * N + j]
As you can see, in order to perform the translation the compiler has to know N, but doesn't really need to know M. This translation formula can easily be generalized for arrays with any number of dimensions. It will involve all sizes of the multi-dimensional array except the first one. This is why every time you declare an array, you are required to specify all sizes except the first one.
As the array in C is purely memory without any meta information about dimensions, the compiler need to know how to apply the row and column index when addressing an element of your matrix.
inputMatrix[i][j] is internally translated to something equivalent to *(inputMatrix + i * NUM + j)
and here you see that NUM is needed.
C doesn't have any specific support for multidimensional arrays. A two-dimensional array such as double inputMatrix[N][M] is just an array of length N whose elements are arrays of length M of doubles.
There are circumstances where you can leave off the number of elements in an array type. This results in an incomplete type — a type whose storage requirements are not known. So you can declare double vector[], which is an array of unspecified size of doubles. However, you can't put objects of incomplete types in an array, because the compiler needs to know the element size when you access elements.
For example, you can write double inputMatrix[][M], which declares an array of unspecified length whose elements are arrays of length M of doubles. The compiler then knows that the address of inputMatrix[i] is i*sizeof(double[M]) bytes beyond the address of inputMatrix[0] (and therefore the address of inputMatrix[i][j] is i*sizeof(double[M])+j*sizeof(double) bytes). Note that it needs to know the value of M; this is why you can't leave off M in the declaration of inputMatrix.
A theoretical consequence of how arrays are laid out is that inputMatrix[i][j] denotes the same address as inputMatrix + M * i + j.¹
A practical consequence of this layout is that for efficient code, you should arrange your arrays so that the dimension that varies most often comes last. For example, if you have a pair of nested loops, you will make better use of the cache with for (i=0; i<N; i++) for (j=0; j<M; j++) ... than with loops nested the other way round. If you need to switch between row access and column access mid-program, it can be beneficial to transpose the matrix (which is better done block by block rather than in columns or in lines).
C89 references: §3.5.4.2 (array types), §3.3.2.1 (array subscript expressions)
C99 references: §6.7.5.2 (array types), §6.5.2.1-3 (array subscript expressions).
¹ Proving that this expression is well-defined is left as an exercise for the reader. Whether inputMatrix[0][M] is a valid way of accessing inputMatrix[1][0] is not so clear, though it would be extremely hard for an implementation to make a difference.
This is because in memory, this is just a contiguous area, a single-dimension array if you will. And to get the real offset of inputMatrix[x][y] the compiler has to calculate (x * elementsPerColumn) + y. So it needs to know elementsPerColumn and that in turn means you need to tell it.
No, there's not. The situation's pretty simple really: what the function receives is really just a single, linear block of memory. Telling it the number of columns tells it how to translate something like block[x][y] into a linear address in the block (i.e., it needs to do something like address = row * column_count + column).
Other people have explained why, but the way to pass a 2D array with unknown dimensions is to pass a pointer. The compiler demotes array parameters to pointers anyway. Just make sure it's clear what you expect in your API docs.
I am passing an Integer array to a function.
Next I want to traverse through the array.
In C and C++ this worked by simply using arrayname.length which gave the number of elements in array. What is the way to find that in Objective-C?
[NSArrayObject length] works for NSArray type but I want it for int[]. Not even [XYZ count] works (an already suggested answer) so I'm looking for another way here.
You can use [XYZ count] to get the length of the array
There isn't anything specific to Objective-C with an array of ints. You would use the same technique as if you were using C.
sz = (sizeof foo) / (sizeof foo[0]);
There is no such thing as array.length in C. An int array in Objective-C is exactly the same as an int array in C. If it's statically defined like int foo[5], then you can do sizeof(foo) to get the size — but only in the same function foo is defined in (to other functions, it's just an int pointer, not an array per se). Otherwise, there is no inherent way to get this information. You need to either pass the size around or use sentinel values (like the terminating '\0' in C strings, which are just arrays of char).
Huh? In C, there's no such thing as "arrayname.length". An array of a primitive type, such as int[], does not have any length information associated with it at runtime.
[array count] this appears to work the easiest in objective-c
this code can be use when the total number of element in array are not known.
main()
{
int a[]={1,2,3,4,5,6,7}
int i;
clrscr();
for (i=0;i<=((sizeof(a)/sizeof(int));i++)
{
puts(a[i]);
}
getch();
}
There is no such thing as arrayname.length in C. That is why so many functions take argument pairs, one argument with the array and one argument with the length of the array. The most obvious case for this is the main function. You can find this function is all your iPhone projects in a file named main.m, it will look something like this:
int main(int argc, char *argv[]) {
NSAutoreleasePool * pool = [[NSAutoreleasePool alloc] init];
int retVal = UIApplicationMain(argc, argv, nil, nil);
[pool release];
return retVal;
}
Notice how the argv agument is a vector, or array of strings, and the argc argument is a count of how many items that are in this array.
You will have to do the same thing for all primitive types, this is the C part of Objective-C. If you stay in the Objective parts using NSArray or subclasses works fine, but requires all elements to be objects.
looks like a job for NSMutableArray. is there a reason why you need to work with a C array? if not, consider NSMutableArray.
i may be wrong (so please correct me if i am), but there's no easy way in C to get the size of int[] at runtime. you would have to do something like create a struct to hold your array and an int where you can keep track of how many items in your array yourself.
even fancier, you can make your struct hold your array and a block. then you can design the block to do the tracking for you.
but this is more a C question, not an objective-c quesiton.
If you want an array of integers, wrap them in NSNumber objects and place them into an NSArray.
What you are asking to do, is not really supported well in C though there is a method on the mac you could try (malloc_size):
determine size of dynamically allocated memory in c
You should try the following approach:
float ptArray[] = {97.8, 92.3, 89.4, 85.7, 81.0, 73.4, 53.0, 42.3, 29.4, 14.1};
int x = sizeof(ptArray)/sizeof(ptArray[0]);
NSLog(#"array count = %d", x);
Output:
array count = 10