Passing an array as argument to a function

Calin Negru

I edited my previous post a bit also

void myFunc(char* somedata)
{
strcpy(somedata, "Good stuff");
return;
}

//...
char myArray[] = "Bad stuff";
myFunc(myArray);
printf("Returned value %s\n", myArray);

Quote:

void myFunc(char* somedata) { strcpy(somedata, "Good stuff"); return; } //... char myArray[] = "Bad stuff"; myFunc(myArray); printf("Returned value %s\n", myArray);

will this function? the array has 9 elements(bad stuff 9 chars.) "good stuff" is 10.

Lost User

Well, of course that breaks it so you may find that your code crashes. You must not overwrite arrays that are passed to you with more data than it can hold. But this conversation is missing some vital information. Maybe you can explain exactly what problem you are trying to solve.

Calin Negru

@Richard Meh I`m really not doing anything worth much attention. My code is stashed with hacks so I`m just trying to fix on it.

"DreamLand Page" on facebook

Lost User

fearless_ wrote:

My code is stashed with hacks

So it all needs a lot of attention.

Calin Negru

Thanks for pointing how things are meant to be done Richard

"DreamLand Page" on facebook

Greg Utas

You can write it like this

void myFunc(char somedata[]) // edited to remove * after char: see Richard's post below

to highlight that it's actually an array rather than just a pointer to a single char. Given that it's an array, its size is often provided to avoid the kind of trampling that you pointed out:

void myFunc(char somedata[], size_t n)

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Calin Negru

thanks Greg for that tip

k5054

Further to what Richard has said, you could also do some defensive programming by passing in the length of the array. eg:

void myFunc(int* somedata, size_t data_len)
{
for(size_t i = 0; i < data_len; ++i)
somedata[i] *= 2;
return;
}

//...
int myArray[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10);
const size_t arrLen = sizeof(myArray)/sizeof(myArray[0]);
myFunc(myArray), arrLen);
// ... etc

This will prevent data overflows, and/or segfaults, when accessing myArray inside the function. Things to note: * we can get the compiler to tell us how many elements in the array using the sizeof(myArray)/sizeof(myArray[0]) construct. This is calculated at compile time and in release mode, in most cases it will be optimized out, so it does not add to your memory usage, if that's a concern. * Using the constant arrLen means that if we change the number of elements in myArray, we don't need to go through the code and find all uses of myArray and make sure we're passing in the right number * If we had used arrLen = sizeof(myArray)/sizeof(int), we need to remember to change the definition of arrLen if we change the type of myArray. Using sizeof(myArray[0] mean that if we change the type of myArray from int[] to double[], for example, we don't need to remember to change the definition of arrLen as well

Keep Calm and Carry On

leon de boer

Under C any pointer is already a pointer to an array it's built into the language along with pointer arithmetic ... you need to just learn that. Literally declare any pointer of anything lets do a float float* p; now you can access it as an array p[0] = 5.0; p[100] = 10.0; It will crash because the pointer isn't really to any memory but it makes the point the pointer is already a pointer to an array There are no exceptions to the rule it doesn't matter if the pointer is to a fundamental type or struct .... so I don't get how you could ever forget that. In the C community the [] use is rare because it's two extra characters to type. It also has implication when declaring variables because it puts that array on the stack not on data memory or constant memory (rodata) if it determines its a constant. So if you get into the habit of using that form you can get some undesirable things happen. Personally you are learning and I would learn to live without it and just learn them as you will most often see them written. The topic is well covered in dummies guide to C but you will note the last statement How to Use Arrays and Functions Together in C Programming - dummies[^]

In vino veritas

Lost User

There is also the _countof() macro, which gives the number of elements in arrays of any simple or composite type.

Lost User

Greg Utas wrote:

char* somedata[]

That's an array of pointers.

leon de boer

good pick didn't even notice the * because that form is so foreign to me.

In vino veritas

leon de boer

Best macro ever added into C standard ... wish more people would use it.

In vino veritas

Lost User

They probably would if it was easier to find; after all it is in the documentation. ;)

k5054

I didn't know about _countof(). Is it an MS only extension? Trying to compile with gcc under linux produces an implicit declaration warning in C and a not declared in this scope in C++

Keep Calm and Carry On

Calin Negru

float* p;
now you can access it as an array
p[0] = 5.0;
p[100] = 10.0;

I understand "p[0] = 5.0;" since that`s the first element, but then "p[100] = 10.0;" don`t you need to alocate first "p = new float[101]"

Lost User

The documentation on MSDN (_countof Macro | Microsoft Docs[^]) does not identify it as MS only.

Greg Utas

Right you are! I forget to remove the *.

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Greg Utas

I made a mistake. See Richard's post below.

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Calin Negru

noticed