更改数组中的 A 字符串时出现 C 分段错误

如何解决更改数组中的 A 字符串时出现 C 分段错误

我有这个指向字符串的指针数组，它在 Main 函数上方定义为 public：

char *Code[]={"MAIN:  add r3,LIST","LOOP: prn #48","lea STR,r6","inc r6","mov r3,K","sub r1,r4","bne END","cmp val1,#-6","bne %END","dec K","jmp %LOOP","END: stop","STR: .string “abcd”","LIST: .data 6,-9",".data -100",".entry K","K: .data 31"};

我写了一个函数，假设查找 A 行中是否有一系列制表符和空格，如果有，它应该更改字符串，以便只有一个空格或制表符：

void SpaceTabRemover()//function to make sure there are no two tabs or spaces following eachother
{
    int i,j=0,k=0; //i=which line we are at,j=which char of the line we ar at,k=placeholder for the last char which equal to space or tab
    for(i=0;i<sizeof(Code)/sizeof(Code[0]);i++)
    {
        while(Code[i][j])
        {
            if ((Code[i][j]==' '||Code[i][j]=='\t')&&(Code[i][j+1]==' '||Code[i][j+1]=='\t'))//checks if there is a sequence of tabs and spaces
            {
                Code[j][k++]=Code[i][j];
            }
            j++;
        }
        Code[j][k]='\0';
    }
}

我认为代码应该可以正常工作，除非我写错了并且我看不到它，

我的问题是当函数确实在 A 行中找到两个空格或制表符并尝试对字符串进行更改时，我收到了一个我以前从未见过的错误：

在 SpaceTabRemover() 在 main.c.112 Code[j][k++]=Code[i][j] 中程序收到信号 SIGSEGV 分段错误

这个错误是什么意思，我该如何解决？

解决方法

arrya Code 的元素是指向字符串文字的指针，不允许修改。

在修改字符串之前，您必须将字符串复制到可修改的缓冲区。

#include <stdlib.h> // for malloc() and exit()
#include <string.h> // for strlen() and strcpy()

void SpaceTabRemover()//function to make sure there are no two tabs or spaces following eachother
{
    int i,j=0,k=0; //i=which line we are at,j=which char of the line we ar at,k=placeholder for the last char which equal to space or tab
    for(i=0;i<sizeof(Code)/sizeof(Code[0]);i++)
    {
        // allocate buffer and copy strings
        char* buffer = malloc(strlen(Code[i]) + 1); // +1 for terminating null-character
        if (buffer == NULL) exit(1); // check if allocation succeeded
        strcpy(buffer,Code[i]); // copy the string
        Code[i] = buffer; // assign the copied string to the element of array

        while(Code[i][j])
        {
            if ((Code[i][j]==' '||Code[i][j]=='\t')&&(Code[i][j+1]==' '||Code[i][j+1]=='\t'))//checks if there is a sequence of tabs and spaces
            {
                Code[j][k++]=Code[i][j];
            }
            j++;
        }
        Code[j][k]='\0';
    }
}