问题描述
AC程序，其代码以下面的简化示例为代表
已在多个编译器上编译后运行了几年
年。然而，对于一个相当新的GCC编译器，它会在指示的指令处导致
分段错误。
简而言之，当写入数组元素时会出现问题
数组指针是由数组索引中的函数分配的。
我的问题：这是不好的C代码，不明智 $ C $ b（注意：这个简单的例子显然不是生产代码。）
感谢您的帮助。
问候，
Charles Sullivan
-------------- ---------------------------
#include< stdio.h>
#include< stdlib.h>
int nextindex（int ** arrayp）
{
static int lastindex = -1;
if（* arrayp == NULL）
* arrayp = calloc（100，sizeof（int））;
返回++ lastindex;
}
int main（无效）
{
int j;
int * array = NULL;
for（j = 0; j< 10; j ++）
array [nextindex（& array）] = 10 * j; / *段错误* /
for（j = 0; j
printf（" array [％d] =％d \ n"，j，array [j]）;
返回0;
}
----- ---------------------------------
A C program with code typified by the following pared-down example
has been running after compilation on numerous compilers for several
years. However with a fairly recent GCC compiler it results in a
segmentation fault at the indicated instruction.
Briefly, the problem occurs when writing to an array element when
the array pointer is allocated by a function in the array index.
My question: Is this bad C code, "unwise" C code, or a compiler bug?
(Note: This simple example is obviously not production code.)
Thanks for your help.
Regards,
Charles Sullivan
-----------------------------------------
#include <stdio.h>
#include <stdlib.h>
int nextindex ( int **arrayp )
{
static int lastindex = -1;
if ( *arrayp == NULL )
*arrayp = calloc(100, sizeof(int));
return ++lastindex;
}
int main ( void )
{
int j;
int *array = NULL;
for ( j = 0; j < 10; j++ )
array[nextindex(&array)] = 10 * j; /* segfaults */
for ( j = 0; j < 5; j++ )
printf("array[%d] = %d\n", j, array[j]);
return 0;
}
--------------------------------------
推荐答案
您不应该检查calloc是否返回有效指针？ （我知道
它与malloc有点不同）
无论如何，* arrayp（或main（）中的数组）的值是多少？ ，就在它出现之前
错了？
-
Bartc
Aren''t you supposed to check whether calloc returns a valid pointer? (I know
it''s a little different to malloc)
What''s the value of *arrayp (or array in main()) anyway, just before it goes
wrong?
--
Bartc
错误的代码。 seg故障线与
*相同（数组+ nextindex（&数组））= 10 * j;
操作数可以按任何顺序进行评估。所以编译器是允许的，例如允许生成将数组值加载到
寄存器中的代码，然后是（可能内联的）nextindex
计算。我的猜测是这就是发生了什么。
Bad code. The seg fault line is the same as
*(array + nextindex(&array)) = 10 * j;
The operands can be evaluated in either order. So the compiler is
allowed to e.g generate code that loads the value of array into a
register, followed by the (potentially inlined) nextindex
computation. My guess is that this is what''s happening.
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