Josh morning Josh morning - 1 year ago 100
C Question

setting the size of array

i am solving a problem. In my code i have two arrays

i am inputing 2 numbers through scanf , and i want the size of arrays to be the count of number 1 and 2 e.g

int x;
int y;
scanf(" %d %d\n",&x,&y);
int indexX[x*y];
int indexY[x*y];

This does not work.
Being new to C im quite lost how to implement it , is the only way dynamicly allocate memory? If no , how can i achieve wanted result ? Or how could i dynamicly alocate the memory for it?
my code throws error

warning: ISO C++ forbids variable length array ‘indexX’ [-Wvla]
warning: ISO C++ forbids variable length array ‘indexY’ [-Wvla]

Answer Source

According to your error messages, you compile with a C++ compiler. Do not! C is not C++ and C++ is not C with classes. They are different languages. As the messages state, C++ does not provide VLAs (see below).

Use a standard-compliant C compiler, or at least a C99 compiler. Variable length arrays (VLAs) were added with the C99 release of the standard as a mandatory feature. C11 relaxed this by making it optional, but most (if not all) compilers which support C99 also implement it in C11 mode.

before defining the array, you should verify scanf really set these two variables by checking the result of scanf. Otherwise you use uninitialised variables, which is undefined behaviour and may (likely - will) result in strange behaviour - at best. Additinally you can also initialise the variables before scanf to 1 (a VLA of size 0 is also undefined behaviour). This way you can check later and are still safe with the VLA definition.

Warning: Most, if not all modern implementations place VLAs on the stack. The size of the stack is normally limited to some 100 bytes (embedded systems) up to some MiB (standard OS like Windows, OS-X, Linux). There is no check if the VLA fits onto the stack, so you should not allocate too large arrays. If you are not sure, better use dynamically allocated memory (malloc& friends).

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