Guys Im passing an array to a function The function defination is

Question

Guys Im passing an array to a function The function defination is

like this:
int my_funcie(int *arr);

It takes an array of integers, and let's say it prints the array's elements

The problem is I want to use a for loop to print elements, and I don't want to hardcode the condition
For example

for(int index = 0; index < array_length; index++)

here, the array_size is the problem
According to this StackOverflow answer

It should work using this piece of code:
int array_length = sizeof(arr) / sizeof(arr[0])

I put this inside my my_funcie function, but it doesn't give me the proper result
For example It's saying the array with 5 integer elemnts (sizeof(arr)) is 8 and the sizeof(arr[0]) is 4
It does 8/4 and gets 2 as the array_length

What am I doing wrong?
The same method works pretty well where I defined the array itself

I'm using C and gcc 9.2.0 btw

0

01.12.2019

3 ответов

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Chandradeep Dey

rewrite the function signature as int my_funcie(size_t size, int arr[size])

0

01.12.2019

Chandradeep Dey

Jens Gustedt's Modern C has multiple paragraphs about the difference between arrays and pointers. you should read them. the book is linked in the pinned message.

0

01.12.2019

Satan · Accepted Answer

Satan

It doesn't work this way if you're getting the array as a function parameter. Go through the ss posted by John it'll explain this in detail. Instead of array you can use vectors. Personally I prefer vectors over arrays plus vectors have their own size() function which will give you an exact count of the number of elements present inside it.

0

01.12.2019

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