（）printf的执行与分割故障 [英] Execution of printf() and Segmentation Fault

问题描述

#include<stdio.h>

int main()
{
    char *name = "Vikram";
    printf("%s",name);
    name[1]='s';
    printf("%s",name);
    return 0;
}

有没有打印输出端子和刚刚获得分段错误。但是，当我在GDB运行它，我获得以下 -

There is no output printed on terminal and just get segmentation fault. But when I run it in GDB, I get following -

Program received signal SIGSEGV, Segmentation fault.
0x0000000000400525 in main () at seg2.c:7
7       name[1]='s';
(gdb) 

这意味着程序接收七号线赛格故障（显然我不能常字符数组写）。那么，为什么行号的printf（）6不执行？

This means program receive SEG fault on 7th line (obviously I can't write on constant char array) . Then why printf() of line number 6 is not executed ?

推荐答案

这是由于流标准输出的缓冲。除非你做 fflush（标准输出）或打印一个新行\\ n输出可被缓冲。

This is due to stream buffering of stdout. Unless you do fflush(stdout) or you print a newline "\n" the output is may be buffered.

在这种情况下，它被冲洗和打印缓冲器之前的段错误。

In this case, it's segfaulting before the buffer is flushed and printed.

您可以改为试试这个：

printf("%s",name);
fflush(stdout);        //  Flush the stream.
name[1]='s';           //  Segfault here (undefined behavior)

或

printf("%s\n",name);   //  Flush the stream with '\n'
name[1]='s';           //  Segfault here (undefined behavior)

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