C: array subscript is below array bounds

Question

void page_kernel_only(int16_t page){

if(mode != KERNEL)
  {
    return;
  }
page = page << 5;
page = page >> 5;
int16_t help = 8192;
help = help & page_table[page];

if(help == 0)
{
    page_table[page] += 8192;
}

}

i get an "array subscript is below array bounds" warning and i dont know why, because a previous part of the code dont give me such a warning

void open_page(int16_t page){
if(mode != KERNEL)
{
return;
}
page = page << 5;
page = page >> 5;
int16_t help = 16384;
help = help & page_table[page];
if(help == 1)
{
    return;
}
page_table[help] += 16384;

}

mmu.c:154:26: warning: array subscript is below array bounds [-Warray-bounds] help = help & page_table[page]; — noaH
– noaH, Commented Jul 13, 2015 at 19:53
Since the variable page is an "int16_t" it's signed. If it's negative the right shift may shift '1' bits in at the most significant bit. I'm not sure what your shifting is actually supposed to accomplish. — Jay
– Jay, Commented Jul 13, 2015 at 19:57
Right shifting a signed integer is implementation defined. Left shifting a signed integer will also generate implementation defined behaviour if the sign is "shifted out". — too honest for this site
– too honest for this site, Commented Jul 13, 2015 at 20:07
page = page << 5; page = page >> 5; What is the purpose of these two statements? If page << 5 overflows, the behavior is undefined. If it doesn't, the net result is that page is unchanged. If you're trying to clear the 5 low-order bits, you'd want to do the right shift followed by the left shift -- but a bitwise & would be a clearer way to do that. — Keith Thompson
– Keith Thompson, Commented Jul 13, 2015 at 20:52

dbush · Accepted Answer · 2015-07-13 20:05:25Z

1

You have page defined as a 16-bit signed int. So if the value is greater than 1024, when you shift left by 5 then shift right by 5 you risk of the value becoming negative, which would generate the warning.

Edit:

If you want to ensure that the top 5 bits of page are 0, you need to do this:

page = page & 0x03FF;

Edit2:

page should be defined as uint16_t instead of int16_t. That should take care of the warnings.

edited Jul 13, 2015 at 20:05

answered Jul 13, 2015 at 19:56

dbush

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4 Comments

noaH Over a year ago

i need it to compile without errors and warnings, is there a workaround to shift back and forth without getting a warning?

too honest for this site Over a year ago

Right shifting a negative value is implementation defined. There is no guarantee for not preserving the sign, nor shifting zeros in.

noaH Over a year ago

int16_t is the guideline for this task, we "shall" learn how to handle those problems and now stackoverflow tought me how to handle them

WhozCraig Over a year ago

@noaH Fyi, per the C11 §6.5.7, p4: "The result of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are filled with zeros. If E1 has an unsigned type, the value of the result is E1 × 2^E2, reduced modulo one more than the maximum value representable in the result type. If E1 has a signed type and nonnegative value, and E1 × 2^E2 is representable in the result type, then that is the resulting value; otherwise, the behavior is undefined.". In short, if page is non-negative, and page << 5 is not representable as an int16_t with value page * 2^5, you've already invoked UB.

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