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I have written a small hello world program, and run strace on its binary, and it listed all system calls which have been called as part of execution of my Hello_world program.

strace ./a.out
execve("./a.out", ["./a.out"], [/* 40 vars */]) = 0
brk(NULL)                               = 0xb7d000

brk(NULL) is used to find the current upper limit of DSS, but my question is, who wanted to know this and why?

edit1: There is no malloc in my program

Ellanti Kishore
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    Possible duplicate: https://unix.stackexchange.com/questions/75638/why-is-brk0-called – SKi Sep 13 '19 at 08:39

2 Answers2

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It is quite likely that, whether you use malloc() or not, the C runtime library prepares the heap. In order to do so, it queries the current size of the DSS and sets up the control structures appropriately in order to be able to act as soon as the first malloc() is called.

glglgl
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I'm writing this as an answer, because I don't have enough reputation to comment. That said, I'd suggest you to look at this and especially this answer.

In a nutshell, when you execute a program, a running process (in your case shell) needs to call fork() to create a new process by duplicating itself. This duplicated process, a "child" process, then calls execve() (which is the first syscall you see listed by strace) which effectively overwrites the text, data, bss, and stack of the calling ("parent") process. Importantly, the memory that might have been dynamically allocated to the parent process before calling fork() is not preserved. Hence, the address returned by brk(NULL) will be different for the two. How and why this information is used by your program is beyond my understanding.

gstukelj
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