printf() 的执行和分段错误 [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;
}

终端上没有打印输出，只是出现分段错误.但是当我在 GDB 中运行它时，我得到了关注 -

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) 

这意味着程序在第 7 行收到 SEG 错误(显然我不能在常量 char 数组上写入).那为什么不执行第 6 行的 printf() 呢?

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 stdout 的流缓冲造成的.除非您执行 fflush(stdout) 或打印换行符 " " 输出可能会被缓冲.

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

在这种情况下，它在缓冲区被刷新和打印之前发生了段错误.

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

你可以试试这个:

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

或:

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

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