二维数组和指针 [英] 2D arrays and pointers
问题描述
嘿,我是一个新手的指针,并在以下code,我想一个2维数组的值存储在一个结构,然后打印出来。不过,我在该行得到一个编译错误:F D-> mychar [I] = NEWPTR [I];
我得到的,而字符* STR
是一样的海峡[]
,的char ** STR
心不是一样海峡[] []
,但我不能找到一个解决方案,使下面的工作。
typedef结构MYSTRUCT {
炭mychar [20] [20];
} mystruct_t;无效函数printValues(字符** NEWPTR){
INT I;
mystruct_t * FD;
对于(I = 0; I&下; 3;我++){
FD-GT&; mychar [I] = NEWPTR [I]
的printf(我的值是%s和结构中的%S \\ n,NEWPTR [I],FD-GT&; mychar [I]);
}
}
INT主(INT ARGC,字符** argv的){
INT I;
字符* ABC [5] = {123,456,789}; 对于(I = 0; I&下; 3;我++){
的printf(我的值是%S \\ n,ABC [I]);
}
函数printValues(ABC);
}
大部分的问题是你的未分配结构的使用。
你用一个指针mystruct_t但从来没有分配它。
对我来说,以下运行:
的#include<&stdio.h中GT;typedef结构MYSTRUCT
{
字符* mychar [20];
} mystruct_t;无效函数printValues(字符** NEWPTR)
{
INT I;
//注意:您在最初的code使用的未分配的指针
mystruct_t FD;
对于(I = 0; I&下; 3;我+ +)
{
fd.mychar [I] = NEWPTR [I]
的printf(我的值是%s和结构中的%S \\ n,NEWPTR [I],fd.mychar [I]);
}
}INT主(INT ARGC,字符** argv的)
{
INT I;
字符* ABC [5] = {123,456,789}; 对于(I = 0; I&下; 3;我+ +)
{
的printf(我的值是%S \\ n,ABC [I]);
}
函数printValues(ABC);
}
Hey All, I am a pointer newbie and in the following code, I am trying to store the values of a 2 D array in a structure and then print them. However, I get a compilation error at the line: fd->mychar[i] = newptr[i];
I get that while char * str
is same as str[]
, char ** str
isnt the same as str[][]
, but I cant find a solution to make the following work.
typedef struct mystruct{
char mychar [20][20];
}mystruct_t;
void printvalues ( char ** newptr){
int i;
mystruct_t * fd;
for (i=0;i<3;i++){
fd->mychar[i] = newptr[i];
printf("My value is %s and in struct %s\n", newptr[i], fd->mychar[i]);
}
}
int main (int argc, char **argv){
int i;
char * abc[5] = {"123", "456", "789"};
for (i=0;i<3;i++){
printf("My value is %s\n", abc[i]);
}
printvalues(abc);
}
Most of the issue was your use of an unallocated structure. You used a pointer to mystruct_t but never allocated it. The following runs for me:
#include <stdio.h>
typedef struct mystruct
{
char* mychar [20];
} mystruct_t;
void printvalues( char** newptr )
{
int i;
// note: you used an unallocated pointer in your original code
mystruct_t fd;
for ( i = 0; i < 3; i++ )
{
fd.mychar[i] = newptr[i];
printf( "My value is %s and in struct %s\n", newptr[i], fd.mychar[i] );
}
}
int main( int argc, char **argv )
{
int i;
char * abc[5] = { "123", "456", "789" };
for ( i = 0; i < 3; i++ )
{
printf( "My value is %s\n", abc[i] );
}
printvalues( abc );
}
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