Pelles C forum
C language => Expert questions => Topic started by: PaoloC13 on May 02, 2020, 01:30:36 AM
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Hi to everybody!
I found myself facing a problem: having a string representing a numeric value, how can I define whether or not it oveflows the range of a specific type? (natively, without using any library!)
The answer I gave myself, lies below. In this example I used the type unsigned int. I would ask if anyone had a simpler and more efficient way.
thanks
#include <stdlib.h>
#include <stdbool.h>
#define _SKIP_LEADING_ZEROES(s) while ( *s == '0' ) s++;
#define ISDIGIT(c) ( c > '/' && c < ':' ) // Only numeric chars.
#define UINT_MAX_DIGITS 10 // undigned int max. 4294967295 (10 digits).
#define ASCII_0 48
#define ASCII_1 49
#define ASCII_2 50
#define ASCII_3 51
#define ASCII_4 52
#define ASCII_5 53
#define ASCII_6 54
#define ASCII_7 55
#define ASCII_8 56
#define ASCII_9 57
inline size_t NumLenA(const char *restrict s)
{
size_t start = (size_t)s;
while( ISDIGIT(*s) ) s++;
return (size_t)s - start;
}
bool UIntOverflowA(const char *p)
{
_SKIP_LEADING_ZEROES(p) // It may be 00000123456789;
size_t len = NumLenA(p);
if( len > UINT_MAX_DIGITS ) return true;
if( len == UINT_MAX_DIGITS )
{
if( p[0] > ASCII_4 ) return true;
if( p[0] == ASCII_4 )
{
if( p[1] > ASCII_2 ) return true;
if( p[1] == ASCII_2 )
{
//if( p[2] > ASCII_9 ) return true;
if( p[2] == ASCII_9 )
{
if( p[3] > ASCII_4 ) return true;
if( p[3] == ASCII_4 )
{
//if( p[4] > ASCII_9 ) return true;
if( p[4] == ASCII_9 )
{
if( p[5] > ASCII_6 ) return true;
if( p[5] == ASCII_6 )
{
if( p[6] > ASCII_7 ) return true;
if( p[6] == ASCII_7 )
{
if( p[7] > ASCII_2 ) return true;
if( p[7] == ASCII_2 )
{
//if( p[8] > ASCII_9 ) return true;
if( p[8] == ASCII_9 )
{
if( p[9] > ASCII_5 ) return true;
}
}
}
}
}
}
}
}
}
}
return false;
}
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Try:
bool UIntOverflow(const char *p)
{
uint32_t v = 0;
while(*p && *p>='0' && *p<='9')
{
uint32_t tmp = v;
v = (v*10) + (uint32_t)(*p-'0');
if (tmp > v)
return false; //Overflow
}
return true;
}
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Hi frankie, thanks for your smart tip. It doesn't seem to work for me, but I tried to fix it:
bool UIntSafeStr(const char *p)
{
uint32_t v = 0;
uint32_t w = 0;
while ( *p >= '0' && *p <= '9' ) // (*p != 0) inclusive.
{
if ( v > 429496729 ) // Test for v *= 10; overflow.
return false;
v *= 10;
w = (unsigned)(*p++ - '0');
if ( UINT_MAX - v < w ) // Test for addition overflow.
return false;
v += w;
}
return true;
}
UIntSafeStr("4294967295"); // Ok!
UIntSafeStr("4294967296"); // Overflow!
Let me know your opinion.
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I should have fixed it at last.
Added increment of string pointer and added a check for shifting overflow.
Try it.
#include <stdio.h>
#include <stdint.h>
#include <stdbool.h>
bool UIntOverflow(const char *p)
{
uint32_t v = 0;
while(*p && *p>='0' && *p<='9')
{
if (v & 0x80000000) //if last bit set
return false; //result will overflow
uint32_t tmp = v;
v = (v * 10) + (uint32_t)(*p++ -'0');
if (tmp > v)
return false; //Overflow
}
return true;
}
int main(int argc, char *argv[])
{
char *str1 = "4294967295";
char *str2 = "4294967296";
printf("%s => %s\n", str1, UIntOverflow(str1) ? "Ok" : "Overflow");
printf("%s => %s\n", str2, UIntOverflow(str2) ? "Ok" : "Overflow");
return 0;
}
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Great! It works properly.
Honestly, I don't understand the need of the "if ( v & 0x80000000 )" check. I was unable to find an input (to the function) that causes this exit.
At the end, it seems to work well even without it, which is your previous solution.
EDIT:
Uh! Found bad case:
4294967295 => Ok
8589934590 => Ok
Press any key to continue...
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Remove the test of last bit set, and try the function for "42949672959"...
Then put back the test before use it in your code! ;)
The check prevents a wraparound that could result on an overflowed value still satisfying the check 'tmp > v'.
Checking last bit reports future overflow before multiplication (just doubling it gives an overflow).
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if you are only interested in a limited case of unsigned int 32, then send the incoming string to a Larger storage variable then compare to int32_max. Check string length less than 11 chars (because if more then alreay too large), then atoll(p) to a long long variable then check if > 4294967295 or int_max. Can generalize ot other sizes if passing size specifier along with string pointer.
Ends up simply
check length
convert to larger storge
compare to max
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I understood that he don't want use any library function and operate natively.
Maybe also the use of larger storage should be avoided?
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Ah, sorry I see that now.... "(natively, without using any library!)"
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Remove the test of last bit set, and try the function for "42949672959"...
Then put back the test before use it in your code! ;)
Yes, it was the case I was looking for and that I had missed. Thanks.
But then, I found other cases that provides an exception, for example:
UIntOverflow("8589934590"); // No owerflow!
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Maybe also the use of larger storage should be avoided?
I'd say, no, I would just like to practice in speed performances without ready-made things that I can't see how they are coded inside.
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Maybe also the use of larger storage should be avoided?
I'd say, no, I would just like to practice in speed performances without ready-made things that I can't see how they are coded inside.
The problem again is to trace eventual overflow at each stage, the check for the last bit wasn't enough.
Last fix could be:
bool UIntOverflow(const char *p)
{
uint32_t v = 0;
while(*p && (*p>='0' && *p<='9'))
{
if (v & 0xe0000000) //if last 3 bits are set multiply
return false; //by 8 or by 2 will overflow
uint32_t tmp = v;
v <<= 3;
if ((v & 0x80000000) && (tmp & 0xc0000000))
return false; //sum with doubled value will overflow
v += tmp << 1; //v*10 done as v*8 + v*2
v += (uint32_t)(*p++ -'0'); //now add next digit
if (tmp > v)
return false; //Overflow
}
return true;
}
Anyway if the size of storage isn't relevant a more simple form is the following:
bool UIntOverflow(const char *p)
{
uint64_t v = 0;
while(*p && (*p>='0' && *p<='9'))
{
v = (v * 10) + (uint64_t)(*p++ -'0');
if (v > 4294967295)
return false; //Overflow
}
return true;
}
Last, for our friends who likes more the assembler, this is a very simple routine in assembler having access to the carry bit.
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Ok frankie, great work! ;)
The last one is the most immediate but also the most performing (if we consider 64 bit processors).
The previous one is a masterpiece of bitwise understanding. I will take the time to go deeper into it. I found it is 1.37 (averagely) slower than the last one.
My fix called UIntSafeStr() I found it is 1.12 (averagely) slower than the last one.
Thanks
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In addition, if you have verified the length of the string before calling this procedure you can speed this up more. There is no need to check the value of v after every digit. Check it once after all digits have been added.
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Thanks Paolo.
Just for fun I made it in assember 32bits:
bool UIntOverflowAsm(const char *p)
{
__asm
{
xor ebx,ebx;
mov edx, p;
repeat:
movsx eax,byte ptr [edx];
cmp eax,0x30; //'0'
jb success;
cmp eax,0x39; //'9'
ja success;
sub eax,0x30; //*p-'0'
shl ebx,1; //ebx = val*2
jc ovflow; //jump if overflow
add eax, ebx; //eax=val*2+actual digit
jc ovflow; //jump if overflow
shl ebx,1; //ebx=val*4
jc ovflow; //jump if overflow
shl ebx,1; //ebx=val*8
jc ovflow; //jump if overflow
add ebx,eax; //ebx=val*8+val*2+actual digit=val*10+actual digit
jc ovflow; //jump if overflow
inc edx; //p++
jmp repeat;
}
ovflow:
return 0;
success:
return 1;
}
For 64 bits the C short version is already optimized.
As suggested computing the length of the string can shorter the test, but, apart the time to compute length, the string must not include non numeric characters. I.e. testing for "0 foo bar foo" will fail.
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For 64 bits the C short version is already optimized.
True it is good, as well as clear, also easy to change for example to int16 checking. But it could be better 'on average' over the set of all possible int32 10 digit inputs. Knowing that only 10 digit inputs need detailed checking ( >10 is overflow, <10 is no overflow) then we also can know that 8/9ths of all 10 digit inputs don't require the full number to be created to determine if an overflow will happen. Only the first digit needs to be checked for a possible quick return. The while loop only needs to run if the first digit is a 4, if it is 5-9 overflow occurs, if 1-3 no overflow. So adding two if's nested before the while loop will speed up the check for 8/9ths of all possible inputs and slow it slightly for 1/9. The 5-9 if first returning overflow, else 1-3 if returning no overflow, else run the While loop to check.
If there was knowledge that most cases would not overflow I'd reverse the 1-3 check and 5-9 check.
I'm probably overthinking it - but hey shelter in place so kind of ,fun ;)
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/* frankie, you are a true source of hight performance code. :) My asm is rusty, but I saved your nice solution. */
Knowing that only 10 digit inputs need detailed checking ( >10 is overflow, <10 is no overflow)
A numeric string can be "000000000123456", it's lenght overflows but the represented value doesn't.
My first horrific and verbose solution at the top of this discussion is about 2.8 time faster than the "uint64_t solution", if a fixed length and well formatted numeric string is assured. But for generic cases I need to use my _SKIP_LEADING_ZEROES(p) macro and then I need to measure the lenght of the numeric with my NumLenA inline function. It is not possible to know the length of a string at runtime without measure it, and this count brings a huge waste of time that makes my cumbersome function no more performing (depends on the input).
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OK I get it - Thanks