如何调试这个C代码

Question

While reading some question on a site I came across below question where ac question needs to be debug

unsigned int a, b, c;
/* a and b are assume to have some values */
c = (a + b) / 2; // <- There is a bug in this st
What is the bug? and how you debug it?

Some of the answer saying it could cause overflow(c=(a+b)/2).but really didn't get how it cause overflow?

Answer 1

a+b may overflow if the sum of a and b is larger than UINT_MAX , the maximum value for an unsigned int . Eg,

unsigned a = 1;
unsigned b = UINT_MAX;

printf("%u\n", (a+b)/2);

prints 0 .

If you want to find the average of two unsigned int s without overflow, do

c = a/2 + b/2 + (a % 2 & b % 2);

(or (a%2 + b%2)/2 , or (a%2 && b%2) , or ((a&1) & (b&1)) , etc.)

Answer 2

If a and/or b are very large then a + b could exceed the maximum size of an unsigned integer (see MAX_UINT in the limits.h file). This would cause an overflow and so the result would be wrong. For example if a and b are both equal to 0x80000000 the result would be 0 in 32-bit arithmetic, rather than expected result 0x80000000.

To solve it you could use something like this instead:

c = a/2 + b/2 + (a % 2 == 1 && b % 2 == 1);

If you know that b is greater than a then you could use this slightly simpler version:

c = a + (b - a) / 2;

Read this article for information about how this bug appeared in binary search algorithms in may popular languages (though it talks about signed int rather than unsigned int ):

• Extra, Extra - Read All About It: Nearly All Binary Searches and Mergesorts are Broken

Answer 3

As other say:

unsigned int a, b, c;
c = (a + b) / 2;

a + b can be not representable in an unsigned int for some value of a and b .

A very similar situation led to a famous bug in the standard Java binary search implementation ( binarySearch function).

See this famous Joshua Blosh blog post in 2006:

"Extra, Extra - Read All About It: Nearly All Binary Searches and Mergesorts are Broken" http://googleresearch.blogspot.com/2006/06/extra-extra-read-all-about-it-nearly.html

Excerpts:

The bug is in this line:

6: int mid =(low + high) / 2;

further down:

So what's the best way to fix the bug? Here's one way:

6: int mid = low + ((high - low) / 2);

Note that in Joshua post, high is >= low and also int objects were used but Java treats signed overflow as wrapping. In C signed integer overflows are undefined behavior and unsigned wraparound.

Answer 4

如果a和b足够高以产生高于unsigned int可表示的最大值的a + b ，则可能导致溢出。

Answer 5

If you only declare a variable in C without initialize it, it gets an unpredictably value . a and b can be some value which is currently at the adress they get in the memory.

You can debug C code in Eclipse CDT
http://www.eclipse.org/cdt/

With this IDE you can program in C anc C++ and it contains the GDB debugger.
http://www.gnu.org/software/gdb/