Bernd Petrovitsch wrote:
On Sun, 2005-08-07 at 15:57 +0200, Thomas Petazzoni wrote:
Hi,
Bob Smith a écrit :
Why is that ? I googled up reasons like minimizing return points,
but is that it ?
Most of the time, 'goto' is used to minimize code duplication in error
handling.
Without 'goto':
int myfunc (void)
{
ret = dosomething1();
if (ret < 0)
return ret;
ret = dosomething2();
if (ret < 0)
{
undosomething1();
return ret;
}
ret = dosomething3();
if (ret < 0)
{
undosomething2();
undosomething1();
return ret;
}
ret = dosomething4();
if (ret < 0)
{
undosomething3();
undosomething2();
undosomething1();
return ret;
}
return 0;
}
With 'goto':
int myfunc(void)
{
ret = dosomething1()
if (ret < 0)
goto ret_dosomething1;
ret = dosomething2()
if (ret < 0)
goto ret_dosomething2;
ret = dosomething3()
if (ret < 0)
goto ret_dosomething3;
ret = dosomething4()
if (ret < 0)
goto ret_dosomething4;
ret_dosomething4:
undosomething3();
ret_dosomething3:
undosomething2();
ret_dosomething2:
undosomething1();
ret_dosomething1:
return ret;
}
With 'if':
int myfunc(void)
{
ret = dosomething1()
if (ret >= 0) {
ret = dosomething2()
if (ret >= 0) {
ret = dosomething3()
if (ret >= 0) {
ret = dosomething4()
if (ret >= 0) {
...
}
}
undosomething3();
}
undosomething2();
}
undosomething1();
return ret;
}
Now remember that the kernel has 8 spaces indentation, not 3 as above.
Not that better, especially if your label names are quite descriptive.
Bernd
Hi Bernd,
IMHO, there is a small problem with the code..
what if ret is > 0? Then the code within the if is executed, only to be
undone after the execution of the if.
for example...
ret = dosomething1()
if (ret > 0) {
/*do stuff */
.......
}
undosomething1()
return ret;
if ret > 0, it will "do stuff" and then undo something1. That IIUC, is
not the desired effect.
Am i right, or is this me misinterpreting the code?
Thanks
Rahul
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