Question

一个简单的程序必须保存为2D数组：

第一行->（'k'，'f'）第二行->（'c'，'d'）

程序是

程序返回分段核心错误的错误。怎么了？

Answer 1

您在线路上遇到问题

**(p+1) = 'f';

您要写入*(p+1)中存储的地址（例如p[1]）的位置，但是您从未初始化p[1]，并且您写入了无效的地址

如果我执行给您以下程序的程序，使用 valgrind 很容易发现这种问题：

==3951== Memcheck, a memory error detector
==3951== Copyright (C) 2002-2017, and GNU GPL'd, by Julian Seward et al.
==3951== Using Valgrind-3.13.0 and LibVEX; rerun with -h for copyright info
==3951== Command: ./a.out
==3951== 
==3951== Use of uninitialised value of size 4
==3951==    at 0x10494: main (c.c:11)
==3951== 
==3951== Invalid write of size 1
==3951==    at 0x10494: main (c.c:11)
==3951==  Address 0x0 is not stack'd, malloc'd or (recently) free'd
==3951== 
==3951== 
==3951== Process terminating with default action of signal 11 (SIGSEGV)
==3951==  Access not within mapped region at address 0x0
==3951==    at 0x10494: main (c.c:11)
==3951==  If you believe this happened as a result of a stack
==3951==  overflow in your program's main thread (unlikely but
==3951==  possible), you can try to increase the size of the
==3951==  main thread stack using the --main-stacksize= flag.
==3951==  The main thread stack size used in this run was 8388608.
==3951== 
==3951== HEAP SUMMARY:
==3951==     in use at exit: 10 bytes in 2 blocks
==3951==   total heap usage: 2 allocs, 0 frees, 10 bytes allocated
==3951== 
==3951== LEAK SUMMARY:
==3951==    definitely lost: 0 bytes in 0 blocks
==3951==    indirectly lost: 0 bytes in 0 blocks
==3951==      possibly lost: 0 bytes in 0 blocks
==3951==    still reachable: 10 bytes in 2 blocks
==3951==         suppressed: 0 bytes in 0 blocks
==3951== Rerun with --leak-check=full to see details of leaked memory
==3951== 
==3951== For counts of detected and suppressed errors, rerun with: -v
==3951== Use --track-origins=yes to see where uninitialised values come from
==3951== ERROR SUMMARY: 2 errors from 2 contexts (suppressed: 6 from 3)

在其中您可以看到对未初始化值的访问及其使用

在C中使用Malloc正确分配2D数组

1 个答案: