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关于多线程写的一个问题，请指教

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关于多线程写的一个问题，请指教

发表于：2007-04-14 11:45:57 楼主

有多个线程，他们都执行一样的操作，即对变量i 进行自增操作，那些线程并没有对变量i的值的一些判断，仅仅就是让i自增，请问这也需要做线程互斥吗？

发表于：2007-04-14 11:54:321楼得分:0

需要，下面的来自《windows核心编程》，可能长了点：）

atomic access: the interlocked family of functions
a big part of thread synchronization has to do with atomic access—a thread 's ability to access a resource with the guarantee that no other thread will access that same resource at the same time. let 's look at a simple example:

// define a global variable.
long g_x = 0;

dword winapi threadfunc1（pvoid pvparam) {
g_x++;
return（0);
}

dword winapi threadfunc2（pvoid pvparam) {
g_x++;
return（0);
}

i 've declared a global variable, g_x, and initialized it to 0. now let 's say that i create two threads: one thread ＥＸＥＣutes threadfunc1, and the other thread ＥＸＥＣutes threadfunc2. the code in these two functions is identical: they both add 1 to the global variable g_x. so when both threads stop running, you might expect to see the value 2 in g_x. but do you? the answer is…maybe. the way the code is written, you can 't tell what g_x will ultimately contain. here 's why. let 's say that the compiler generates the following code for the line that increments g_x by 1:

mov eax, [g_x] ; move the value in g_x into a register.
inc eax ; increment the value in the register.
mov [g_x], eax ; store the new value back in g_x.

both threads are unlikely to ＥＸＥＣute this code at exactly the same time. so if one thread ＥＸＥＣutes this code followed by another thread, here is what effectively ＥＸＥＣutes:

mov eax, [g_x] ; thread 1: move 0 into a register.
inc eax ; thread 1: increment the register to 1.
mov [g_x], eax ; thread 1: store 1 back in g_x.

mov eax, [g_x] ; thread 2: move 1 into a register.
inc eax ; thread 2: increment the register to 2.
mov [g_x], eax ; thread 2: store 2 back in g_x.

after both threads are done incrementing g_x, the value in g_x is 2. this is great and is exactly what we expect: take zero （0), increment it by 1 twice, and the answer is 2. beautiful. but wait—windows is a preemptive, multithreaded environment. so a thread can be switched away from at any time and another thread might continue ＥＸＥＣuting at any time. so the code above might not ＥＸＥＣute exactly as i 've written it. instead, it might ＥＸＥＣute as follows:

mov eax, [g_x] ; thread 1: move 0 into a register.
inc eax ; thread 1: increment the register to 1.

mov eax, [g_x] ; thread 2: move 0 into a register.
inc eax ; thread 2: increment the register to 1.
mov [g_x], eax ; thread 2: store 1 back in g_x.

mov [g_x], eax ; thread 1: store 1 back in g_x.

if the code ＥＸＥＣutes this way, the final value in g_x is 1—not 2 as you expect! this is pretty scary, especially since you have so little control over the scheduler. in fact, if you have 100 threads ＥＸＥＣuting similar thread functions, after all of them exit, the value in g_x might still be 1! obviously, software developers can 't work in an environment like this. we expect that incrementing 0 twice results in 2 all the time. also, let 's not forget that the results might be different depending on how the compiler generates codes, what cpu is ＥＸＥＣuting the code, and how many cpus are installed in the host computer. this is how the environment works, and there is nothing we can do about that. but windows does offer some functions that, when used correctly, guarantee the outcome of our application 's code.

to solve the problem above, we need something simple. we need a way to guarantee that the incrementing of the value is done atomically—that is, without interruption. the interlocked family of functions provides the solution we need. the interlocked functions are awesome and underused by most software developers, even though they are incredibly helpful and easy to understand. all of the functions manipulate a value atomically. take a look at interlockedexchangeadd:

long interlockedexchangeadd（
plong pladdend,
long lincrement);

what could be simpler? you call this function, passing the address of a long variable and indicating by how much to increment this value. but this function guarantees that the adding of the value is accomplished atomically. so we can rewrite the code presented earlier as follows:

// define a global variable.
long g_x = 0;

dword winapi threadfunc1（pvoid pvparam) {
interlockedexchangeadd（&g_x, 1);
return（0);
}

dword winapi threadfunc2（pvoid pvparam) {
interlockedexchangeadd（&g_x, 1);
return（0);
}

by making this small change, g_x is incremented atomically and therefore you are guaranteed that the final value in g_x will be 2. don 't you feel better already? note that all the threads should attempt to modify the shared long variable by calling these functions; no thread should ever attempt to modify the shared variable by using simple c statements:

// the long variable shared by many threads
long g_x;

// incorrect way to increment the long
g_x++;

// correct way to increment the long
interlockedexchangeadd（&g_x, 1);

how do the interlocked functions work? the answer depends on the cpu platform that you 're running on. for the x86 family of cpus, interlocked functions assert a hardware signal on the bus that prevents another cpu from accessing the same memory address. on the alpha platform, the interlocked functions do something like this:

发表于：2007-04-14 11:55:072楼得分:0

turn on a special bit flag in the cpu and note the memory address being accessed.

read the value from memory into a register.

modify the register.

if the special bit flag in the cpu is off, go to step 2. otherwise, the special bit flag is still on and the register 's value is stored back into memory.

you might wonder how the special cpu bit flag is ever turned off by the time step 4 is ＥＸＥＣuted. here 's the answer: if another cpu in the system attempts to modify the same memory address, it can turn off our cpu 's special bit flag, causing the interlocked function to loop back to step 2.

you need not understand exactly how the interlocked functions work. what 's important to know is that they guarantee that a value will be modified atomically, no matter how the compiler generates code and no matter how many cpus are installed in the host machine. you must also ensure that the variable addresses that you pass to these functions are properly aligned or the functions might fail. （i 'll discuss data alignment in chapter 13.)

another important thing to know about the interlocked functions is that they ＥＸＥＣute extremely quickly. a call to an interlocked function usually causes just a few cpu cycles （usually less than 50) to ＥＸＥＣute, and there is no transition from user mode to kernel mode （which usually requires more than 1000 cycles to ＥＸＥＣute).

of course, you can use interlockedexchangeadd to subtract a value—you simply pass a negative value for the second parameter. interlockedexchangeadd returns the original value that was in *pladdend.

here are two more interlocked functions:

long interlockedexchange（
plong pltarget,
long lvalue);

pvoid interlockedexchangepointer（
pvoid* ppvtarget,
pvoid pvvalue);

interlockedexchange and interlockedexchangepointer atomically replace the current value whose address is passed in the first parameter with a value passed in the second parameter. for a 32-bit application, both functions replace a 32bit value with another 32-bit value. but for a 64-bit application, interlockedexchange replaces a 32-bit value while interlockedexchangepointer replaces a 64-bit value. both functions return the original value. interlockedexchange is extremely useful when you implement a spinlock:

// global variable indicating whether a shared resource is in use or not
bool g_fresourceinuse = false;

发表于：2007-04-14 11:55:183楼得分:0

void func1（) {
// wait to access the resource.
while （interlockedexchange （&g_fresourceinuse, true) == true)
sleep（0);

// access the resource.

// we no longer need to access the resource.
interlockedexchange（&g_fresourceinuse, false);
}

the while loop spins repeatedly, changing the value in g_fresourceinuse to true and checking its previous value to see if it was true. if the value was previously false, the resource was not in use but the calling thread just set it to in-use and exits the loop. if the previous value was true, the resource was in use by another thread and the while loop continues to spin.

if another thread were to ＥＸＥＣute similar code, it would spin in its while loop until the g_fresourceinuse was changed back to false. the call to interlockedexchange at the end of the function shows how g_fresourceinuse should be set back to false.

you must take extreme care when using this technique because a spinlock wastes cpu time. the cpu must constantly compare two values until one "magically " changes due to another thread. also, this code assumes that all threads using the spinlock run at the same priority level. you might also want to disable thread priority boosting （call setprocesspriorityboost or setthreadpriorityboost) for threads that ＥＸＥＣute spinlocks.

in addition, you should ensure that the lock variable and the data that the lock protects are maintained in different cache lines （discussed later in this chapter). if the lock variable and data share the same cache line, a cpu using the resource will contend with any cpus attempting access of the resource. this hurts performance.

you should avoid using spinlocks on single-cpu machines. if a thread is spinning, it 's wasting precious cpu time, which prevents the other thread from changing the value. my use of sleep in the while loop above improves this situation somewhat. if you use sleep, you might want to sleep a random amount of time, and each time the request to access the resource is denied, you might want to increase the sleep time even more. this prevents threads from simply wasting cpu time. depending on your situation, it might be better to remove the call to sleep altogether. or you might want to replace it with a call to switchtothread （not available on windows 98). i hate to say it, but trial and error might be your best approach.

spinlocks assume that the protected resource is always accessed for short periods of time. this makes it more efficient to spin and then transition to kernel mode and wait. many developers spin some number of times （say 4000), and if access to the resource is still denied, the thread transitions to kernel mode, where it waits （consuming no cpu time) until the resource becomes available. this is how critical sections are implemented.

spinlocks are useful on multiprocessor machines because one thread can spin while the other thread runs on another cpu. however, even in this scenario, you must be careful. you do not want a thread to spin for a long time, or you 'll waste more cpu time. we 'll discuss spinlocks further later in this chapter. also, the section titled "implementing a critical section: the optex " in chapter 10 shows how to use spinlocks.

here are the last two interlocked functions:

pvoid interlockedcompareexchange（
plong pldestination,
long lexchange,
long lcomparand);

pvoid interlockedcompareexchangepointer（
pvoid* ppvdestination,
pvoid pvexchange,
pvoid pvcomparand);

these two functions perform an atomic test and set operation: for a 32-bit application, both functions operate on 32-bit values, but in a 64-bit application, interlockedcompareexchange operates on 32-bit values while interlockedcompareexchangepointer operates on 64-bit values. in pseudocode, here is what happens:

long interlockedcompareexchange（plong pldestination,
long lexchange, long lcomparand) {

long lret = *pldestination; // original value

if （*pldestination == lcomparand)
*pldestination = lexchange;
return（lret);
}

the function compares the current value （pointed to by the pldestination parameter) with the value passed in the lcomparand parameter. if the values are the same, *pldestination is changed to the value of the lexchange parameter. if what is in *pldestination doesn 't match the value of lcomparand, *pldestination is not changed. the function returns the original value in *pldestination. remember that all of these operations are performed as one atomic unit of ＥＸＥＣution.

there is no interlocked function that simply reads a value （without changing it) because no such function is necessary. if a thread simply attempts to read the contents of a value that is always modified with an interlocked function, the value read is always a good value. you don 't know if you 'll read the original value or the updated value, but you know that it will be one of them. for most applications, this is sufficient. in addition, the interlocked functions might be used by threads in multiple processes when you 're synchronizing access to a value that is in a shared memory section such as a memory-mapped file. （chapter 9 includes a few sample applications that show how to properly use the interlocked functions.)

windows offers a few other interlocked functions, but the functions i 've described do everything that the other functions do and more. here are two other functions:

long interlockedincrement（plong pladdend);

long interlockeddecrement（plong pladdend);

interlockedexchangeadd replaces both of these older functions. the new function can add or subtract any value; the old functions are limited to adding or subtracting 1.

发表于：2007-04-15 19:16:354楼得分:0

用原子操作即可，如上面说的interlockedincrement（)

发表于：2007-04-15 21:21:295楼得分:0

如果变量i是线程内的局部变量，不需要，否则就需要，总的来说，如果一个线程读写的是一个本线程内的局部变量，不需要互斥，而对于本线程外的变量，则一定要互斥，除非所有的线程都是对这个变量读

发表于：2007-04-16 09:52:366楼得分:0

用interlockedincrement,不需要互斥,但是要对变量做保护

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