libiop | C++ library for IOP-based zkSNARKs | Cryptography library

The comment by Ctx:

This is not possible; sbrk() is used by the libc memory allocator internally, so many libc functions use it implicitly. Either do not use libc at all, or do your memory allocations with malloc() or mmap().

is correct. You can only use sbrk if you don't use malloc, and you can't know if you use malloc because unless documented otherwise by the implementation, anything in libc might do so.

If your libc supports replacing malloc (note: most do) then you can write a full malloc replacement (this includes all malloc-family functions, not just malloc itself) that either don't use sbrk, or that cooperate with your use of it, and then you're free to use it. However otherwise, you simply can't use sbrk.

Note also that in some environments, including frequently with position-independent executables (PIE), sbrk will have little or no memory to work with and will frequently fail after only a few allocations or after none at all, because of running into memory mapped for something else. The whole concept of sbrk is backwards and should not be used at all in modern code. For some allocator strategies, it may make sense as one choice of backend, with mmap or something else also used. But it should never be the sole way your allocator obtains new memory, and in my opinion it's not really useful to support it at all. mmap is better in almost all ways.