CBMC在以下几行中检测到可能的未签名加法溢出:
l = (t + *b)&(0xffffffffL);
c += (l < t);
我同意第一行中可能会发生溢出，但是我正在照顾下一行中CBMC无法查看的进位。
如果发生溢出，请设置进位1。因此，由于我知道这一点，因此我希望我的代码能够正常工作，我希望继续进行验证过程。
那么，我告诉CBMC如何忽略此错误并继续前进呢？
最佳答案
TL; DR 这取决于变量的实际类型。在所有情况下，CBMC都会检测到可能导致不确定行为的实际错误。这意味着，您应该修复您的代码，而不是禁用CBMC中的消息。
完整答案:
通常:据我所知，CBMC不允许排除特定属性(另一方面，您可以使用--property标志仅检查一个特定属性)。如果您需要正式的答案/意见或提出功能要求，建议您在CProver Support group中发布。
(当然，可以使用__CPROVER_assume来使CBMC排除导致错误的痕迹，但这是非常非常非常糟糕的主意，因为这可能会使其他问题无法实现。)
变体1:我假设您的代码看起来像这样(与此相关:请，请发布自包含的示例并准确说明问题所在，很难猜出这些事情)
long nondet_long(void);
void main(void) {
long l = 0;
int c = 0;
long t = nondet_long();
long s = nondet_long();
long *b = &s;
l = (t + *b) & (0xffffffffL);
c += (l < t);
}
而你正在运行
cbmc --signed-overflow-check test.c
giving an output similar to the following one?
CBMC version 5.1 64-bit x86_64 macos
Parsing test.c
Converting
Type-checking test
Generating GOTO Program
Adding CPROVER library
Function Pointer Removal
Partial Inlining
Generic Property Instrumentation
Starting Bounded Model Checking
size of program expression: 41 steps
simple slicing removed 3 assignments
Generated 2 VCC(s), 2 remaining after simplification
Passing problem to propositional reduction
converting SSA
Running propositional reduction
Post-processing
Solving with MiniSAT 2.2.0 with simplifier
792 variables, 2302 clauses
SAT checker: negated claim is SATISFIABLE, i.e., does not hold
Runtime decision procedure: 0.006s
Building error trace
Counterexample:
State 17 file test.c line 4 function main thread 0
----------------------------------------------------
l=0 (0000000000000000000000000000000000000000000000000000000000000000)
State 18 file test.c line 4 function main thread 0
----------------------------------------------------
l=0 (0000000000000000000000000000000000000000000000000000000000000000)
State 19 file test.c line 5 function main thread 0
----------------------------------------------------
c=0 (00000000000000000000000000000000)
State 20 file test.c line 5 function main thread 0
----------------------------------------------------
c=0 (00000000000000000000000000000000)
State 21 file test.c line 6 function main thread 0
----------------------------------------------------
t=0 (0000000000000000000000000000000000000000000000000000000000000000)
State 22 file test.c line 6 function main thread 0
----------------------------------------------------
t=-9223372036854775808 (1000000000000000000000000000000000000000000000000000000000000000)
State 23 file test.c line 7 function main thread 0
----------------------------------------------------
s=0 (0000000000000000000000000000000000000000000000000000000000000000)
State 24 file test.c line 7 function main thread 0
----------------------------------------------------
s=-9223372036854775807 (1000000000000000000000000000000000000000000000000000000000000001)
State 25 file test.c line 8 function main thread 0
----------------------------------------------------
b=((long int *)NULL) (0000000000000000000000000000000000000000000000000000000000000000)
State 26 file test.c line 8 function main thread 0
----------------------------------------------------
b=&s!0@1 (0000001000000000000000000000000000000000000000000000000000000000)
Violated property:
file test.c line 10 function main
arithmetic overflow on signed + in t + *b
!overflow("+", signed long int, t, *b)
VERIFICATION FAILED
I do not think you should disable this property check, even if you could. The reason for this is, as you say, that this addition can overflow, and, integer overflow is undefined behaviour in C, or, as this answer to the question How to check integer overflow in C? nicely puts it:
See also Integer overflow and undefined behavior and How disastrous is integer overflow in C++?.
Thus, this is an actual bug and CBMC has a good reason for telling you about it. What you actually should do is adapt your code so that there are no potential overflows! The answer mentioned above suggests something like (remember to include limits.h):
if ((*b > 0 && t > LONG_MAX - *b)
|| (*b < 0 && LONG_MIN < *b && t < LONG_MIN - *b)
|| (*b==LONG_MIN && t < 0))
{
/* Overflow will occur, need to do maths in a more elaborate, but safe way! */
/* ... */
}
else
{
/* No overflow, addition is safe! */
l = (t + *b) & (0xffffffffL);
/* ... */
}
变体2:在这里，我假设您的代码如下所示:
unsigned int nondet_uint(void);
void main(void) {
unsigned int l = 0;
unsigned int c = 0;
unsigned int t = nondet_uint();
unsigned int s = nondet_uint();
unsigned int *b = &s;
l = (t + *b) & (0xffffffffL);
c += (l < t);
}
而你正在运行
cbmc --unsigned-overflow-check test.c
giving an output similar to the following one?
CBMC version 5.1 64-bit x86_64 macos
Parsing test.c
Converting
Type-checking test
Generating GOTO Program
Adding CPROVER library
Function Pointer Removal
Partial Inlining
Generic Property Instrumentation
Starting Bounded Model Checking
size of program expression: 42 steps
simple slicing removed 3 assignments
Generated 3 VCC(s), 3 remaining after simplification
Passing problem to propositional reduction
converting SSA
Running propositional reduction
Post-processing
Solving with MiniSAT 2.2.0 with simplifier
519 variables, 1306 clauses
SAT checker: negated claim is SATISFIABLE, i.e., does not hold
Runtime decision procedure: 0.01s
Building error trace
Counterexample:
State 17 file test.c line 4 function main thread 0
----------------------------------------------------
l=0 (00000000000000000000000000000000)
State 18 file test.c line 4 function main thread 0
----------------------------------------------------
l=0 (00000000000000000000000000000000)
State 19 file test.c line 5 function main thread 0
----------------------------------------------------
c=0 (00000000000000000000000000000000)
State 20 file test.c line 5 function main thread 0
----------------------------------------------------
c=0 (00000000000000000000000000000000)
State 21 file test.c line 6 function main thread 0
----------------------------------------------------
t=0 (00000000000000000000000000000000)
State 22 file test.c line 6 function main thread 0
----------------------------------------------------
t=4187126263 (11111001100100100111100111110111)
State 23 file test.c line 7 function main thread 0
----------------------------------------------------
s=0 (00000000000000000000000000000000)
State 24 file test.c line 7 function main thread 0
----------------------------------------------------
s=3329066504 (11000110011011011000011000001000)
State 25 file test.c line 8 function main thread 0
----------------------------------------------------
b=((unsigned int *)NULL) (0000000000000000000000000000000000000000000000000000000000000000)
State 26 file test.c line 8 function main thread 0
----------------------------------------------------
b=&s!0@1 (0000001000000000000000000000000000000000000000000000000000000000)
Violated property:
file test.c line 10 function main
arithmetic overflow on unsigned + in t + *b
!overflow("+", unsigned int, t, *b)
VERIFICATION FAILED
Again, this is an actual bug and CBMC has a good reason for telling you about it. This one could be fixed by
l = ((unsigned long)t + (unsigned long)*b) & (0xffffffffL);
c += (l < t);
这使
CBMC version 5.1 64-bit x86_64 macos Parsing test.c Converting Type-checking test Generating GOTO Program Adding CPROVER library Function Pointer Removal Partial Inlining Generic Property Instrumentation Starting Bounded Model Checking size of program expression: 42 steps simple slicing removed 3 assignments Generated 3 VCC(s), 3 remaining after simplification Passing problem to propositional reduction converting SSA Running propositional reduction Post-processing Solving with MiniSAT 2.2.0 with simplifier 542 variables, 1561 clauses SAT checker inconsistent: negated claim is UNSATISFIABLE, i.e., holds Runtime decision procedure: 0.002s VERIFICATION SUCCESSFUL
Variant 3: If things are as the previous one, but you have signed int instead of unsigned int, things get a bit more complicated. Here, assuming you use (written in a slightly more elaborate way to better see what is going on)
int nondet_int(void);
void main(void) {
int l = 0;
int c = 0;
int t = nondet_int();
int s = nondet_int();
long longt = (long)t;
long longs = (long)s;
long temp1 = longt + longs;
long temp2 = temp1 & (0xffffffffL);
l = temp2;
c += (l < t);
}
并运行
cbmc --signed-overflow-check test.c
you will get
CBMC version 5.1 64-bit x86_64 macos Parsing test.c Converting Type-checking test Generating GOTO Program Adding CPROVER library Function Pointer Removal Partial Inlining Generic Property Instrumentation Starting Bounded Model Checking size of program expression: 48 steps simple slicing removed 3 assignments Generated 3 VCC(s), 3 remaining after simplification Passing problem to propositional reduction converting SSA Running propositional reduction Post-processing Solving with MiniSAT 2.2.0 with simplifier 872 variables, 2430 clauses SAT checker: negated claim is SATISFIABLE, i.e., does not hold Runtime decision procedure: 0.008s Building error trace Counterexample: State 17 file test.c line 4 function main thread 0 ---------------------------------------------------- l=0 (00000000000000000000000000000000) State 18 file test.c line 4 function main thread 0 ---------------------------------------------------- l=0 (00000000000000000000000000000000) State 19 file test.c line 5 function main thread 0 ---------------------------------------------------- c=0 (00000000000000000000000000000000) State 20 file test.c line 5 function main thread 0 ---------------------------------------------------- c=0 (00000000000000000000000000000000) State 21 file test.c line 6 function main thread 0 ---------------------------------------------------- t=0 (00000000000000000000000000000000) State 22 file test.c line 6 function main thread 0 ---------------------------------------------------- t=-2147483648 (10000000000000000000000000000000) State 23 file test.c line 7 function main thread 0 ---------------------------------------------------- s=0 (00000000000000000000000000000000) State 24 file test.c line 7 function main thread 0 ---------------------------------------------------- s=1 (00000000000000000000000000000001) State 25 file test.c line 9 function main thread 0 ---------------------------------------------------- longt=0 (0000000000000000000000000000000000000000000000000000000000000000) State 26 file test.c line 9 function main thread 0 ---------------------------------------------------- longt=-2147483648 (1111111111111111111111111111111110000000000000000000000000000000) State 27 file test.c line 10 function main thread 0 ---------------------------------------------------- longs=0 (0000000000000000000000000000000000000000000000000000000000000000) State 28 file test.c line 10 function main thread 0 ---------------------------------------------------- longs=1 (0000000000000000000000000000000000000000000000000000000000000001) State 29 file test.c line 11 function main thread 0 ---------------------------------------------------- temp1=0 (0000000000000000000000000000000000000000000000000000000000000000) State 31 file test.c line 11 function main thread 0 ---------------------------------------------------- temp1=-2147483647 (1111111111111111111111111111111110000000000000000000000000000001) State 32 file test.c line 12 function main thread 0 ---------------------------------------------------- temp2=0 (0000000000000000000000000000000000000000000000000000000000000000) State 33 file test.c line 12 function main thread 0 ---------------------------------------------------- temp2=2147483649 (0000000000000000000000000000000010000000000000000000000000000001) Violated property: file test.c line 14 function main arithmetic overflow on signed type conversion in (signed int)temp2 temp2 = -2147483648l VERIFICATION FAILED
Or, written more concisely, if you have
t == -2147483648 (0b10000000000000000000000000000000)
s == 1 (0b00000000000000000000000000000001)
然后
temp2 == 2147483649 (0b0000000000000000000000000000000010000000000000000000000000000001)
尝试将其转换为
signed int是麻烦的，因为它超出了范围(另请参见Does cast between signed and unsigned int maintain exact bit pattern of variable in memory?)。如您所见，这个反例也是一个实际的错误，同样，CBMC正确地告诉了您有关此错误的信息。特别是这意味着您的掩码/数学无法按预期方式工作(您的掩码将负数变为超出范围的正数)，您需要修复代码，以使结果在必要的范围内。 (可能值得认真考虑一下您实际想要做什么，以确保获得正确的结果。)
关于c - 绕过CBMC检测到的未签名加法溢出，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/31065205/