Question

我的c函数'strcpy'有问题，我无法弄清楚。

它涉及复制到像Argv（但实际上不是Argv）的char * []。我可以复制出结构，但不能复制。但只有我最初在一次性中声明整个Argv结构。

我假设char *[]是char*的数组。

以下是该问题的简单演示程序：

#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。那些不可写。尽管出于历史原因他们的类型为char*，但您应始终将其视为char const *。

malloc缓冲区的char空间，或者使用数组数组。

Answer 2

正如dasblinkenlight在上面评论中发布的链接

char * p = "xyz"; is different from 

char p[] = "xyz";

第一个是不可变的，第二个是可变的。

Why do I get a segmentation fault when writing to a string initialized with "char *s" but not "char s[]"?

使用char * []（ARGV结构）进行strcpy崩溃

2 个答案: