c - strcpy 因 char *[] 崩溃(ARGV 结构)

Question

这个问题已经有答案了:

Why do I get a segmentation fault when writing to a "char *s" initialized with a string literal, but not "char s[]"? (19 个回答)

已关闭 9 年前。

我无法解决 c 函数“strcpy”的问题。

它涉及像 Argv 一样复制到 char *[](但实际上不是 Argv)。我可以从结构中复制出来，但不能复制进去。但前提是我最初一次性声明了整个 Argv 结构。

我假设 char *[]是 char* 的数组.

这是一个简单的演示程序:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

char FN[]="BBBB";   

char *TL[]={
 "a0                ",
 "a1                ",
 "a2                ",
 "a3                ",
 "a4                ",
 "a5                "};

 char BN[]="c1                    ";

 char N0[]="N0                    ";
 char N1[]="N1                    ";
 char N2[]="N2                    ";
 char N3[]="N3                    ";
 char N4[]="N4                    ";
 char N5[]="N5                    ";

 char* TD[6]; 


 int main ( int argc, char *argv[] )
 {

    // FN is a pointer to an array of chars
    // BN is the same

    //TL is an array of pointers (that each point to an array of chars)

    //TL[1] is thus a pointer to an array of chars

    //TL is the same structure as Argv

    //TD is the same structure as Argv  (but built up from parts)
    //  but is spread out across the globals and the main func.
    //  thus less easy to read and understand then TL.

    //TL[i], TD[i], and BN are initially allocated significantly larger than FN
    //  to remove the worry of overruns.

    //copy "a1                \0" into the space held by "c1   "
    strcpy(BN,TL[1]); //works

    //copy "BBBB\0" into the space held by "c1   "
    strcpy(BN,FN); //works

    TD[0]=N0;
    TD[1]=N1;
    TD[2]=N2;
    TD[3]=N3;
    TD[4]=N4;
    TD[5]=N5;

    //copy "BBBB\0" into the space held by "a1   "
    strcpy(TD[1],FN); //works

    //copy "BBBB\0" into the space held by "a1   "
    //strcpy(TL[1],FN); //dies



}

Answer 1

您的char指针指向字符串文字。这些是不可写的。即使由于历史原因它们的类型是 char*，您也应该始终将它们视为 char const *。

为 char 缓冲区使用 malloc 空间，或者使用数组的数组。