[英]segmentation fault using char 2d array in c
I nead to create a 2d char arry looking like that: 我需要创建一个看起来像这样的2d char arry:
----
|..|
|..|
----
to do that i wrote a function numed Build_Screen looking like that 为此,我编写了一个编号为Build_Screen的函数,看起来像那样
#include<stdio.h>
#include"build.h"
void Build_Screen(char **c, int w, int k)
{
int i=0, j=0;
for(; i<w;i++)
{
j=0;
for(; j<k;j++)
{
if (i==0||i==w-1)
c[i][j]='-';
else
{
if(j==0||j==k-1)
c[i][j]='|';
else
c[i][j]='.';
}
}
}
}
when i tried to run that like that 当我试图那样跑
char screen[5][5];
Build_Screen(screen, 5, 5);
it caused segmentation foult. 它导致分割。 how do i fix that? 我该如何解决?
The problem is that your function taking 问题是您的功能正在使用
char **
is not the same as char[5][5]
so just change 与char[5][5]
所以只需更改
void Build_Screen(char **c, int w, int k)
with 与
void Build_Screen(char c[5][5], int w, int k)
You have to pass the array, not a pointer
to char pointer
, you can see here , it's explained. 你必须通过数组,而不是一个pointer
到char pointer
,你可以看到在这里 ,它的解释。
If you don't know the size before hand, then use malloc this way 如果您事先不知道大小,请以这种方式使用malloc
void Build_Screen(char **c, int w, int k)
{
int i, j;
for(i=0;i<w;i++)
{
for(j=0;j<k;j++)
{
if (i == 0 || i == w-1)
c[i][j] = '-';
else
{
if(j==0||j==k-1)
c[i][j]='|';
else
c[i][j]='.';
}
}
}
}
int main(int argc, char **argv)
{
char **screen;
int i, rowCount, columnCount;
rowCount = 5;
columnCount = 5;
screen = malloc(rowCount * sizeof(char *));
if (screen == NULL)
return -1;
for (i = 0 ; i < rowCount ; i++)
{
screen[i] = malloc(columnCount);
if (screen[i] == NULL)
{
i -= 1;
for( ; i >= 0 ; i--)
free(screen[i]);
free(screen);
return -1;
}
}
Build_Screen(screen, rowCount, columnCount);
/* finished using it */
for (i = 0 ; i < rowCount ; i++)
free(screen[i]);
free(screen);
return 0;
}
notice that I am assuming that you will print the rows without using printf
or similar functions, if you are going to, then just add a terminating '\\0'
at the end of the strings. 请注意,我假设您将在不使用printf
或类似函数的情况下打印行,如果要这样做,则只需在字符串的末尾添加一个终止符'\\0'
。
Actually, when you state : 实际上,当您声明:
char screen[5][5];
What is happening ? 怎么了 ? It allocates 5*5*sizeof (char)
contiguous bytes on the stack, pointed by screen[0]
. 它在堆栈上分配5*5*sizeof (char)
个连续字节,由screen[0]
指向。 Thus, screen[0][1]
is strictly equivalent to *(screen[0] + 0*5 + 1);
因此, screen[0][1]
严格等于*(screen[0] + 0*5 + 1);
Your problem can be easily resolved by changing your function's signature to 您可以通过将函数的签名更改为来轻松解决问题
void Build_Screen(char c[5][5], int w, int k);
I guess you want to have a function that works with any size of maps. 我想您想拥有一个适用于任何尺寸地图的功能。 This won't happen with genuine 2D array in C, except in C99. 在C中使用正版2D阵列将不会发生这种情况,但在C99中除外。 More details available here . 此处有更多详细信息。
A char[5][5] is a 2D array and not a char**. char [5] [5]是2D数组,而不是 char **。 It is a contiguous zone of 25 chars : 这是一个25个字符的连续区域:
00 01 02 03 04 10 11 12 13 14 20 ...
A char **
is a pointer to a pointer to a char : char **
是指向char **
的指针的指针 :
0 -> 00 01 02 03 04
1 -> 10 11 12 13 14
2 -> 20 ...
3 -> 30 ...
4 -> 40 ...
You must declare it that way : 您必须这样声明:
char ascreen[5][5];
char *screen[5];
for(i=0; i<5; i++) screen[i] = &(ascreen[i][0]);
Build_Screen(screen, 5, 5);
Edit: here is a full program : 编辑:这是一个完整的程序:
#include<stdio.h>
void Build_Screen(char **c, int w, int k)
{
int i=0, j=0;
for(; i<w;i++)
{
j=0;
for(; j<k;j++)
{
if (i==0||i==w-1)
c[i][j]='-';
else
{
if(j==0||j==k-1)
c[i][j]='|';
else
c[i][j]='.';
}
}
}
}
int main() {
int i;
char ascreen[5][5];
char *screen[5];
for(i=0; i<5; i++) screen[i] = &(ascreen[i][0]);
Build_Screen(screen, 5, 5);
for(i=0; i<5; i++) {
printf("%.5s\n", screen[i]); /* or printf("%.5s\n", ascreen[i]); */
}
return 0;
}
and the output is : 输出为:
-----
|...|
|...|
|...|
-----
You can also access your array as a one dimensional array. 您还可以将数组作为一维数组访问。 The following code was sucessfully tested locally: 以下代码已在本地成功测试:
void Build_Screen(char* c, int w, int k)
{
int i = 0, j = 0;
for (; i<w; i++)
{
j = 0;
for (; j<k; j++)
{
if (i == 0 || i == w - 1)
c[i*w+j] = '-';
else
{
if (j == 0 || j == k - 1)
c[i*w + j] = '|';
else
c[i*w + j] = '.';
}
}
}
}
int main()
{
char screen[5][5];
Build_Screen((char*)screen, 5, 5);
}
As @iharob wrote, your function's signature is not working as you expect. 正如@iharob所写,函数的签名未按预期工作。 You should use a single pointer char *c and index the array like this or something similar: 您应该使用单个指针char * c并像这样或类似的方式索引数组:
c[i+w*j]
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