新手问题：为什么这样做？ [英] Newbie question: Why does this work?

问题描述

大家好！

这肯定是一个全新的问题，但为什么

以下代码会导致段错误？


void insertion_sort（int a []，int length）

{

int i;

for（i = 0 ; i< length; i ++）

{

int j，v = a [i];

for（j = i - 1 ; j> = 0; j--）

{

if（a [j]< = v）{

cout< ;< a [j]<< " < =" << v<<结束;

休息;

}

a [j + 1] = a [j];

}

a [j + 1] = v;

}

}


恕我直言，应该发生段错误在第9行，如第一次通过

第一个for-loop i = o ergo j为-1，导致[j]引发

段错误。 />

提前谢谢，请原谅我提出这样一个愚蠢的问题

解决方案

博士******************* @ gmail.com 说：

大家好！

这肯定是一个全新的问题，但为什么不是

以下代码导致段错误？


void insertion_sort（int a []，int length）

{

int i;

for（i = 0; i< length; i ++）

{

int j，v = a [i];

for（j = i - 1; j> = 0; j--）

{

if（a [j]< = v）{


这并没有因为，第一次通过，i = 0，所以j变为i - 1 =

-1，这不是> = 0，所以循环体被完全跳过。

cout<< a [j]<< " < =" << v<< ENDL;

仅供记录，这是一个C新闻组，而不是C ++新闻组。


< snip>


-

Richard Heathfield< http://www.cpax.org.uk>

电子邮件：-http：/ /万维网。 + rjh @

谷歌用户：< http://www.cpax.org.uk/prg/writings/googly.php>

Usenet是一个奇怪的放置" - dmr 1999年7月29日

Ph ******************* @ gmail.com 写道：

嗨全部！

这肯定是一个全新的问题，但是为什么不跟随代码的

导致段错误？


void insertion_sort（int a []，int length）

{

int i;

for（i = 0; i<长度; i ++）

{

int j，v = a [i];

for（j = i - 1; j>如果（a [j] <= v）{

cout<< a [j]<< " < =" << v<<结束;

休息;

}

a [j + 1] = a [j];

}

a [j + 1] = v;

}

}


恕我直言，应该发生段错误在第9行，如第一次通过

第一个for-loop i = o ergo j为-1，导致[j]引发

段错误。 />

在此先感谢，请原谅我提出这样一个愚蠢的问题

第二个for循环只会在它''时输入s条件为真，

ie，仅当j为> = 0时才会发生这种情况。这不会发生在外循环的第一次

迭代之后。在外部
循环的第二次迭代期间，j将初始化为0，因此将执行内部for循环

下的代码。

文章< 64 ********************************* * @ a23g2000hsc。 googlegroups.com>，
Ph ******* ************ @ gmail.com < Ph ******************* @ gmail.comwrote：

>这肯定是一个全新的问题，但为什么以下代码不会导致段错误？

> void insertion_sort（int a []，int length）
$

int i;

for（i = 0; i< length; i ++）

{
int j，v = a [i];
for（j = i - 1; j> = 0; j--）
{
if（a [j]< = v）{
cout<< a [j]<< " < =" << v<< ENDL;

上面的行不会编译，因为你没有定义名为''cout''的变量

，也没有定义名为''的变量ENDL。此外，左移'/ b $ ba指针的值，或指针的值，不是在C中定义的

操作。也许在复制程序中用于发布，

你意外地从另一个窗口粘贴在一行中，你b $ b有一些东西与其他编程语言有关，比如

Rogaine？

>休息;
}
a [j + 1] = a [j];
}
a [j + 1] = v;

}
}

>恕我直言，segfault应该在第9行发生，就像第一次通过第一个for-loop一样i = o ergo j为-1，导致[j]引发一个
段错误。

当i = 0时，你启动for（j = i - 1; j> = 0; j--）循环，

然后j -will-被初始化为-1，但循环条件j> = 0

将在通过循环进行任何行程之前进行评估，并将

被发现为false，因此循环不会被执行。当j不是至少为0时，j> = 0

保护循环不被执行。

-

Man生活只是一种玩笑，

梦想，阴影，泡沫，空气，最好的蒸汽。

- George Walter Thornbury

Hi all!
This is most certainly a total newbie question, but why doesn''t the
following code cause a segfault?

void insertion_sort(int a[], int length)
{
int i;
for (i=0; i < length; i++)
{
int j, v = a[i];
for (j = i - 1; j >= 0; j--)
{
if (a[j] <= v) {
cout << a[j] << " <= " << v << endl;
break;
}
a[j + 1] = a[j];
}
a[j + 1] = v;
}
}

IMHO the segfault should occur at line 9 as on the first pass through
the first for-loop i = o ergo j is -1 causing a[j] to provoke a
segfault.

Thanks in advance and excuse me for asking such a dumb question

解决方案

Ph*******************@gmail.com said:

Hi all!
This is most certainly a total newbie question, but why doesn''t the
following code cause a segfault?

void insertion_sort(int a[], int length)
{
int i;
for (i=0; i < length; i++)
{
int j, v = a[i];
for (j = i - 1; j >= 0; j--)
{
if (a[j] <= v) {

This isn''t broken because, first time through, i = 0, so j becomes i - 1 =
-1, which is not >= 0, so the loop body is skipped completely.

cout << a[j] << " <= " << v << endl;

Just for the record, this is a C newsgroup, not a C++ newsgroup.

<snip>

--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

Ph*******************@gmail.com wrote:

Hi all!
This is most certainly a total newbie question, but why doesn''t the
following code cause a segfault?

void insertion_sort(int a[], int length)
{
int i;
for (i=0; i < length; i++)
{
int j, v = a[i];
for (j = i - 1; j >= 0; j--)
{
if (a[j] <= v) {
cout << a[j] << " <= " << v << endl;
break;
}
a[j + 1] = a[j];
}
a[j + 1] = v;
}
}

IMHO the segfault should occur at line 9 as on the first pass through
the first for-loop i = o ergo j is -1 causing a[j] to provoke a
segfault.

Thanks in advance and excuse me for asking such a dumb question

The second for loop will be entered only when it''s condition is true,
i.e., only when j is >= 0. This will not happen after the first
iteration of the outer loop. During the second iteration of the outer
loop, j will be initialised to 0 so the code under the inner for loop
will be executed.

In article <64**********************************@a23g2000hsc. googlegroups.com>,
Ph*******************@gmail.com <Ph*******************@gmail.comwrote:

>This is most certainly a total newbie question, but why doesn''t the
following code cause a segfault?

>void insertion_sort(int a[], int length)
{
int i;
for (i=0; i < length; i++)
{
int j, v = a[i];
for (j = i - 1; j >= 0; j--)
{
if (a[j] <= v) {
cout << a[j] << " <= " << v << endl;

The above line will not compile, as you have not defined a variable
named ''cout'', nor a variable named ''endl''. Furthermore, left-shifting
a pointer by a value, or a value by a pointer, is not a defined
operation in C. Perhaps in copying out the program for the posting,
you accidently pasted in a line from another window in which you
had something dealing with some other programming language such as
Rogaine ?

> break;
}
a[j + 1] = a[j];
}
a[j + 1] = v;
}
}

>IMHO the segfault should occur at line 9 as on the first pass through
the first for-loop i = o ergo j is -1 causing a[j] to provoke a
segfault.

When i = 0, and you start the for (j = i - 1; j >= 0; j--) loop,
then j -will- be initialized to -1, but the loop condition j >= 0
will be evaluated before any trips are taken through the loop, and will
be found to be false, so the loop will not be executed. The j >= 0
protects the loop from executing when j is not at least 0.
--
"Man''s life is but a jest,
A dream, a shadow, bubble, air, a vapor at the best."
-- George Walter Thornbury

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