[英]How to store a string in a multidimensional array in C
I'm new at C and I'm trying to do an exercise which asks to insert some strings and then store them.我是 C 的新手,我正在尝试做一个要求插入一些字符串然后存储它们的练习。 First it requests a multidimensional array where we have for every row of the array a string, and then as an array of pointers.
首先,它请求一个多维数组,其中数组的每一行都有一个字符串,然后是一个指针数组。 Here's the code for the first part.
这是第一部分的代码。 I don't know how to store into an array some strings that are not already written.
我不知道如何将一些尚未写入的字符串存储到数组中。
For the second one I have no idea since I've never done exercises with pointers before.对于第二个我不知道,因为我以前从未用指针做过练习。
#include <stdio.h>
int main(){
int n; //number of strings
int x; //number of characters per string
printf("How many strings do you want to insert?");
scanf("%d", &n);
if ((n >= 1) && (n <= 20)){
printf("How many characters per string?");
scanf("%d", &x);
char str[x];
if (x <= 10){
for(int i = 0; i < n; i++){
printf("Insert a string:");
scanf("%s", str);
for(int j = 0; j < x; j++){
char arr[j];
arr[j] = str[x];
printf("%s", arr);
}
}
}
else {
printf("Error:the number of characters must be < 10");
}
}
else {
printf("Error: the number must be < 20");
}
return 0;
}
... requests a multidimensional array where we have for every row of the array a string, and then as an array of pointers.
...请求一个多维数组,其中数组的每一行都有一个字符串,然后是一个指针数组。
After getting the qualified number of strings, allocate an array of pointers to char
.在获得合格的字符串数量后,分配一个指向
char
的指针数组。
if ((n >= 1) && (n <= 20)){
char **string_list = calloc(n, sizeof *string_list);
assert(string_list); // or other error checking
(Notice no type in = calloc(n, sizeof *string_list);
. Easier to code right, review and maintain.) (注意没有 type in
= calloc(n, sizeof *string_list);
。更容易正确编码、查看和维护。)
Read the strings in a working temp buffer.读取工作临时缓冲区中的字符串。 As "How many characters per string?"
作为“每个字符串有多少个字符?” likely means the number of characters not including the null character , our
str[]
needs a +1 in size.可能意味着不包括null 字符的字符数,我们的
str[]
需要 +1 大小。
// char str[x]; // too small
char str[x+1];
Yet we know x <= 10
and can use a fixed buffer size and limit input length然而我们知道
x <= 10
并且可以使用固定的缓冲区大小并限制输入长度
for(int i = 0; i < n; i++){
char str[10+1];
printf("Insert a string:");
scanf("%10s", str); // Notice the 10 - a width limit
// TBD check if scanf() returned 1 and if str is longer than x
Now allocate a copy of the str
现在分配
str
的副本
string_list[j] = strdup(str);
assert(string_list[j]); // or other error checking
}
Later, when done with string_list[]
, clean-up and free allocations.稍后,当使用
string_list[]
完成时,清理和释放分配。
for (int i=0; i<n; i++) {
free(string_list[i]);
}
free(string_list);
What is weak about this:这有什么弱点:
It uses scanf()
rather than fgets()
and then parses, has minimal error checking, does not take in strings with spaces, does not handle over-long input, strdup()
is not standard -yet , etc.它使用
scanf()
而不是fgets()
然后解析,具有最少的错误检查,不接受带空格的字符串,不处理过长的输入, strdup()
还不是标准的,等等。
So the above is a baby step.所以上面是一个婴儿步骤。 Better code would handle the weak issues.
更好的代码将处理薄弱的问题。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.