Why is my constant array stored on the stack instead of a text section?

Question

I am implementing an encryption algorithm (for educational purposes) and I noticed something strange. Part of the algorithm uses s-boxes for substitution, so I allocated const arrays for use as a lookup table as follows:

 const unsigned char s0_lookup[4][4]={{1,0,3,2}, {3,2,1,0}, {0,2,1,3}, {3,1,3,2}}; const unsigned char s1_lookup[4][4]={{0,1,2,3}, {2,0,1,3}, {3,0,1,0}, {2,1,0,3}}; 

Since arrays use the const qualifier, I thought they should be stored in the text area, not on the stack. However, if I parse the compiler output, I see the following:

 0000000000000893 <s_des_sbox>: 893: 55 push %rbp 894: 48 89 e5 mov %rsp,%rbp 897: 48 89 7d c8 mov %rdi,-0x38(%rbp) 89b: c6 45 dd 00 movb $0x0,-0x23(%rbp) 89f: c6 45 e0 01 movb $0x1,-0x20(%rbp) 8a3: c6 45 e1 00 movb $0x0,-0x1f(%rbp) 8a7: c6 45 e2 03 movb $0x3,-0x1e(%rbp) 8ab: c6 45 e3 02 movb $0x2,-0x1d(%rbp) 8af: c6 45 e4 03 movb $0x3,-0x1c(%rbp) 8b3: c6 45 e5 02 movb $0x2,-0x1b(%rbp) 8b7: c6 45 e6 01 movb $0x1,-0x1a(%rbp) 8bb: c6 45 e7 00 movb $0x0,-0x19(%rbp) 8bf: c6 45 e8 00 movb $0x0,-0x18(%rbp) 8c3: c6 45 e9 02 movb $0x2,-0x17(%rbp) 8c7: c6 45 ea 01 movb $0x1,-0x16(%rbp) 8cb: c6 45 eb 03 movb $0x3,-0x15(%rbp) 8cf: c6 45 ec 03 movb $0x3,-0x14(%rbp) 8d3: c6 45 ed 01 movb $0x1,-0x13(%rbp) 8d7: c6 45 ee 03 movb $0x3,-0x12(%rbp) 8db: c6 45 ef 02 movb $0x2,-0x11(%rbp) 8df: c6 45 f0 00 movb $0x0,-0x10(%rbp) 8e3: c6 45 f1 01 movb $0x1,-0xf(%rbp) 8e7: c6 45 f2 02 movb $0x2,-0xe(%rbp) 8eb: c6 45 f3 03 movb $0x3,-0xd(%rbp) 8ef: c6 45 f4 02 movb $0x2,-0xc(%rbp) 8f3: c6 45 f5 00 movb $0x0,-0xb(%rbp) 8f7: c6 45 f6 01 movb $0x1,-0xa(%rbp) 8fb: c6 45 f7 03 movb $0x3,-0x9(%rbp) 8ff: c6 45 f8 03 movb $0x3,-0x8(%rbp) 903: c6 45 f9 00 movb $0x0,-0x7(%rbp) 907: c6 45 fa 01 movb $0x1,-0x6(%rbp) 90b: c6 45 fb 00 movb $0x0,-0x5(%rbp) 90f: c6 45 fc 02 movb $0x2,-0x4(%rbp) 913: c6 45 fd 01 movb $0x1,-0x3(%rbp) 917: c6 45 fe 00 movb $0x0,-0x2(%rbp) 91b: c6 45 ff 03 movb $0x3,-0x1(%rbp) 

The code moves literal type constants to populate an empty array on the stack! It seems terribly inefficient to me when the whole array can simply be saved as a constant. Why does my code do this?

Answer 1

As it is declared in a function and not static, it is usually allocated on the stack. Since C allows recursion, each new call to your function will receive a new fresh copy of the array, filled at runtime.

To initialize it only once during build, you must make it static:

 static const unsigned char s0_lookup[4][4]={{1,0,3,2}, {3,2,1,0}, {0,2,1,3}, {3,1,3,2}}; 

As const is declared, optimization can use the as-if rule and compile it, as you wrote static const ... , but nothing forces the compiler to execute it.