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When I print the contents of the running process, I got result like this

00400000-00401000 r-xp 00000000 08:01 137910 /home/user/process/a.out 00600000-00601000 rw-p 00000000 08:01 137910 /home/user/process/a.out 02434000-02455000 rw-p 00000000 00:00 0 [heap]

Why the a.out is loaded at two memory regions? Is it the same part or different parts. One has got r-xp and other has got rw-p. Eventhough I have allocated a size of integer using malloc in the program, the size of the heap is shown more than that. Why is it so?

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The r-xp region at 0x400000 is your text (code) section. It's readable, and executable, but not writable.

The rw-p region at 0x600000 is your data section. It's readable and writable, but not executable.

readelf -S a.out will show you the sections in your executable, and where they are going to be loaded into memory (first through a section-to-segment mapping.)

Your experiment about mallocing 4 bytes is insufficient, because that's not how memory management works. When you call malloc, your libc implementation is going to carve out a small piece out of the large pool that it is maintaining. (This is just your process doing this - the OS is not immediately involved.) When you have depleted that pool, it will use the brk (or mmap) system call to ask the kernel to give it more memory.

If you malloc a much larger amount of memory, you will probably see the heap grow. Also, you can run strace on your executable, and see when it actually makes the brk or memmap system calls.

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