[英]search array for string
我將如何實現這一點?我試圖將c字符串數組與單個字符串進行比較,如果沒有匹配則將其附加到2d數組。
char*mprt,uni[100][16];
mprt = &uni[0][0];
for (int s = 0;s <= 99;s++)
{
for (int z = 0;z <= 15;z++)
{
if (strcmp(mprt++, string1) != 0)
{
uni[s][z] = string1[z];
}
}
}
在for循環中,您需要復制整個字符串以附加它,
用此替換線,
strcpy(uni[s], string1[z]);
考慮string1[z]
是char指針數組的元素。
編輯:
不確定這是否是你想要做的,但你最終將所有元素設置為string1
char string1[] = "String";
char uni[100][16] = {};
for (int s = 0; s < 100; s++)
{
if (strcmp(uni[s], string1) != 0)
{
strcpy(uni[s], string1);
}
}
或者這個,沒有strcpy()
char string1[] = "String";
char uni[100][16] = {};
for (int s = 0; s < 100; s++)
{
for (int r = 0; r < sizeof(string1); r++)
{
uni[s][r] = string1[r];
}
}
好的......從你的評論我現在得到你想要做的。 你想把它變成一個函數,這樣你就可以向它提供單詞,但它應該讓你指向正確的方向。
請注意,您可以使用char[][]
,但這樣您的字符串可以是任意長度,因為我們在將它們放入列表時動態分配它們。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
/* space for 100 strings */
char **uni = calloc(100, sizeof(char*));
char **i;
/* Put one word in the list for test */
*uni = calloc(5, sizeof(char*));
strncpy(*uni, "this", 5);
/* here's the string we're going to search for */
char * str2 = "that";
/* go through the first dimension looking for the string
note we have to check that we don't exceed our list size */
for (i = uni; *i != NULL && i < uni+100; i++)
{
/* if we find it, break */
if (strcmp(*i,str2) == 0)
break;
}
/* if we didn't find the string, *i will be null
* or we will have hit the end of our first dimension */
if (i == uni + 100)
{
printf("No more space!\n");
}
else if (*i == NULL)
{
/* allocate space for our string */
*i = calloc(strlen(str2) + 1, sizeof(char));
/* copy our new string into the list */
strncpy(*i, str2, strlen(str2) + 1);
}
/* output to confirm it worked */
for (i = uni; *i != NULL && i < uni+100; i++)
printf("%s\n",*i);
}
為完整char[][]
, char[][]
版本:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char uni[100][16];
int i,j;
/* init our arrays */
for (i=0;i<100;i++)
for (j=0;j<16;j++)
uni[i][j] = '\0';
/* Put one word in the list for test */
strncpy(uni[0], "this",15);
/* here's the string we're going to search for */
char * str2 = "that";
/* go through the first dimension looking for the string */
for (i = 0; uni[i][0] != '\0' && i < 100; i++)
{
/* if we find it, break */
if (strcmp(uni[i],str2) == 0)
break;
}
/* if we didn't find the string, uni[i][0] will be '\0'
* or we will have hit the end of our first dimension */
if (i == 100)
{
printf("No more space!\n");
}
else if (uni[i][0] == '\0')
{
/* copy our new string into the array */
strncpy(uni[i], str2, 15);
}
/* output to confirm it worked */
for (i = 0; uni[i][0] != '\0' && i < 100; i++)
printf("%s\n",uni[i]);
}
編輯以解釋以下評論中的C指針和數組:
在C中,數組降級為指針。 當你第一次開始時,這實際上真的很混亂。
如果我有char myArray[10]
並且我想將它傳遞給一個帶有char *
參數的函數,我可以使用&myArray[0]
或myArray
。 當你離開索引時,它會降級為指向數組中第一個元素的指針。
在像你這樣的多維數組中, &uni[5][0]
== uni[5]
- 兩者都指向第一個中索引為5的第二個維度中的第一個元素。 它降級為char*
指向列表中第6個單詞的開頭。
要附加到2D數組的末尾,您需要使用動態內存分配
const int row_max = 100, col_max = 16;
char** uni = NULL;
char searchString[col_max] = "xyz";
int currentLength = 0;
uni = (char**) malloc (row_max * sizeof(char*)); //TODO:error handling code to be added
for (int row = 0; row < row_max; row++)
{
uni[row] = (char*)malloc(col_max * sizeof(char));//TODO:error handling code to be added
currentLength = row;
}
for (int row = 0; row < row_max; row++) //fill array uni with data here
{
uni[row] = "abc";
}
for (int row = 0; row < row_max; row++)
{
for (int col = 0; col < col_max; col++)
{
if (strcmp(&uni[row][col], searchString) != 0 )
{//string not found
uni = (char**)realloc(uni, (currentLength + 1) * sizeof(char*));//TODO:error handling code to be added
uni[currentLength + 1] = (char*)malloc(col_max);//TODO:error handling code to be added
currentLength++;
strcpy(uni[currentLength],searchString); //append at end of 2D array
goto stop;
}
}
}
stop:for(int row = 0; row <= currentLength; row ++)free(uni [row]); 自由(UNI);
return 0;
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