如何在二维数组中存储多个字符串？

Question

我想编写一个 C 程序，在运行时存储一堆句子，并在程序结束时打印出来。这是我写的：

#include<string.h>
#include<stdio.h>

int main(){
    int counter = 0;
    char state[100][200];
    char stroo[3] = "yes";
    sprintf(state[counter], "%s", stroo);
    counter++;
    char poo[2] = "44";
    sprintf(state[counter], "%s", poo);
    counter++;
    for (int i=0; i<counter; i++) {
        printf("%s\n", state[i]);
    }
    return 0;
}

对于输出，第一行打印出 "yes"，这正是我所期望的。然而，第二行打印“44yes”。为什么打印 "yes" 以及“44”？

Answer 1

那是因为你的字符串没有正确地以 null 结尾，所以你运行在 sprintf 处进入了未定义的行为。这是因为缓冲区太小，无法容纳空终止符。例如这里：

char stroo[3] = "yes";

您需要存储四个字符，'y'、'e'、's' 和空终止符。改成

char stroo[4] = "yes";

同样 char poo[2] 需要 char poo[3]。

参见 here（强调我的）：

Successive bytes of the string literal or wide characters of the wide string literal, including the terminating null byte/character, initialize the elements of the array [...] If the size of the array is known, it may be one less than the size of the string literal, in which case the terminating null character is ignored:

如果您希望尺寸自动足够大以完全适合字符串，您可以省略尺寸说明：

char stroo[] = "yes";

当然，对于这个例子，您甚至不需要将字符串复制到堆栈中：

char *stroo = "yes";

Answer 2

#define STRINGS 5
#define STRING_SIZE 50

char arr[STRINGS][STRING_SIZE] =
{ "Hi",
  "Hello world",
  "Good morning",
  "ASDF",
  "QWERTY"
};

for (int i=0; i<STRINGS; i++) {
    printf("%s\n", arr[i]);
}

如何在二维数组中存储多个字符串？

How to store multiple strings in a 2-D array?

c

arrays

string

c11