[英]How to find an element in an array in C
I am trying to find the location of an element in the array. 我试图在数组中找到元素的位置。 I have tried to use this code i generated
我试图使用我生成的这段代码
for(i=0;i<10;i++)
{
if (strcmp(temp[0],varptr[i])==0) j=i;
}
varptr is a pointer which points to array var[11][10] and it is by the definition *varptr[11][10]. varptr是一个指向数组var [11] [10]的指针,根据定义* varptr [11] [10]。 I have assigned strings to var[i] and i want to get the "i" number of my element NOT THE ADRESS.
我已将字符串分配给var [i],我想获取我的元素“ THE ADRESS”的“ i”号。
Thanks for any comment. 感谢您的任何评论。
EDit: temp is also a pointer which points to the string that i want to check. EDit:temp也是一个指向我要检查的字符串的指针。 Also i am using the 2D array for keeping variable names and their address.
我也使用2D数组来保留变量名及其地址。 So yes i want to keep it inside a 2D array.
所以是的,我想将其保留在2D数组中。 The question is this code is not working at all, it does not assigns i to j, so i wonder where is the problem with this idea?
问题是此代码根本不起作用,它没有将i分配给j,所以我想知道这个想法在哪里出问题? writing a "break" does not change if the code works or not, it just optimizes the code a little.
编写“中断”不会改变代码是否起作用,只是稍微优化了代码。
Full Code: 完整代码:
#include <stdio.h>
#include <string.h>
#include <ctype.h>
double atof(char*);
int main(void)
{
char in[100], *temp[10],var[11][10],*varptr[11][10];
int i,j, n = 0,fullval=0;
double val[11];
strcpy(var[11], "ans");
for(i=0;i<11;i++)
{
for(j=0;j<10;j++) varptr[i][j]=&var[i][j];
}
START:
printf("Enter the expression: ");
fflush(stdout);
for(i=0;i<10;i++) temp[i]=NULL;
if (fgets(in, sizeof in, stdin) != NULL)
{
temp[0] = strtok(in, " ");
if (temp[0] != NULL)
{
for (n = 1; n < 10 && (temp[n] = strtok(NULL," ")) != NULL; n++)
;
}
if (*temp[0]=="quit")
{
goto FINISH;}
if (isdigit(*temp[0]))
{
if (*temp[1]=='+') val[0] = atof(temp[0])+atof(temp[2]);
else if (*temp[1]=='-') val[0] = atof(temp[0])-atof(temp[2]);
else if (*temp[1]=='*') val[0] = atof(temp[0])*atof(temp[2]);
else if (*temp[1]=='/') val[0] = atof(temp[0])/atof(temp[2]);
printf("%s = %f\n",var[11],val[0]);
goto START;
}
else
if (temp[1]==NULL) //asking the value of a variable
{
for(i=0;i<10;i++)
{
if (strcmp(temp[0],varptr[i])==0) j=i;
}
printf("%s = %d\n",var[j],val[j]);
goto START;
}
if (*temp[1]==61)
{
strcpy(var[fullval], temp[0]);
if ((temp[3])!=NULL)
{
}
val[fullval]=atof(temp[2]);
printf("%s = %f\n",var[fullval],val[fullval]);
fullval++;
goto START;
}
if (*temp[1]!=61)
{
}
}
getch();
FINISH:
return 0;
}
j=-1;
for(i=0;i<10;i++)
{
if (strcmp(temp[0],varptr[i])==0) {j=i;break;}
}//strcmp is not safe, try use strncmp
One comment: you can exit the loop as soon as you find your string. 一条评论:找到字符串即可立即退出循环。
#define NOT_FOUND (-1)
int j = NOT_FOUND;
int i;
for (i = 0 ; i < 11 && j == NOT_FOUND; i++)
{
if (strncmp(temp,var[i], 10) == 0) // Nick D's comment
{
j = i;
}
}
Another comment: 另一条评论:
I was unable to understand how varptr and var relate to each other (please show definitions). 我无法理解varptr和var如何相互关联(请显示定义)。 I have used var in the above on the assumption it is defined:
我已经在上面基于定义的假设中使用了var:
char var[11][10];
also 也
char temp[10];
int i;
int found = 0;
for (i = 0 ; i < 11 ; i++)
{
if (strcmp(temp,var[i]) == 0)
{
found = 1;
break;
}
}
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