这是我编写的用于复制字符串常量的程序。
程序运行时会崩溃。为什么会这样?
#include <stdio.h>
char *alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char c;
char *l;
main(){
while((c = *alpha++)!='\0')
*l++ = *alpha;
printf("%s\n",l);
}
要在 C 中复制字符串,可以使用 strcpy。这是一个例子:
#include <stdio.h>
#include <string.h>
const char * my_str = "Content";
char * my_copy;
my_copy = malloc(sizeof(char) * (strlen(my_str) + 1));
strcpy(my_copy,my_str);
如果您想避免意外的缓冲区溢出,请使用strncpy
而不是strcpy
. 例如:
const char * my_str = "Content";
const size_t len_my_str = strlen(my_str) + 1;
char * my_copy = malloc(len_my_str);
strncpy(my_copy, my_str, len_my_str);
要执行此类手动复制:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char* orig_str = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char* ptr = orig_str;
// Memory layout for orig_str:
// ------------------------------------------------------------------------
// |0|1|2|3|4|5|6|7|8|9|10|11|12|13|14|15|16|17|18|19|20|21|22|23|24|25|26| --> indices
// ------------------------------------------------------------------------
// |A|B|C|D|E|F|G|H|I|J|K |L |M |N |O |P |Q |R |S |T |U |V |W |X |Y |Z |\0| --> data
// ------------------------------------------------------------------------
int orig_str_size = 0;
char* bkup_copy = NULL;
// Count the number of characters in the original string
while (*ptr++ != '\0')
orig_str_size++;
printf("Size of the original string: %d\n", orig_str_size);
/* Dynamically allocate space for the backup copy */
// Why orig_str_size plus 1? We add +1 to account for the mandatory
// '\0' at the end of the string.
bkup_copy = (char*) malloc((orig_str_size+1) * sizeof(char));
// Place the '\0' character at the end of the backup string.
bkup_copy[orig_str_size] = '\0';
// Current memory layout for bkup_copy:
// ------------------------------------------------------------------------
// |0|1|2|3|4|5|6|7|8|9|10|11|12|13|14|15|16|17|18|19|20|21|22|23|24|25|26| --> indices
// ------------------------------------------------------------------------
// | | | | | | | | | | | | | | | | | | | | | | | | | | |\0| --> data
// ------------------------------------------------------------------------
/* Finally, copy the characters from one string to the other */
// Remember to reset the helper pointer so it points to the beginning
// of the original string!
ptr = &orig_str[0];
int idx = 0;
while (*ptr != '\0')
bkup_copy[idx++] = *ptr++;
printf("Original String: %s\n", orig_str);
printf("Backup String: %s\n", bkup_copy);
return 0;
}
您需要为l
. 目前它指向内存中的一个随机点,如果您尝试写入该点,操作系统可能会关闭(AKA 崩溃)您的应用程序。如果您希望您的代码按原样工作,则为l
with分配一些空间malloc()
或创建l
为具有足够空间以容纳"ABCDEFGHIJKLMNOPQRSTUVWXYZ"
加上 NULL 终止符的字符数组。
有关指针的入门知识,请参见http://cslibrary.stanford.edu/106/。
复制字符串“常量/文字/指针”
char *str = "some string thats not malloc'd";
char *tmp = NULL;
int i = 0;
for (i = 0; i < 6; i++) {
tmp = &str[i];
}
printf("%s\n", tmp);
反转
char *str = "some stupid string";
char *tmp, *ptr = NULL;
ptr = str;
while (*str) { ++str; }
int len = str - ptr;
int i = 0;
for (i = len; i > 11; i--) {
tmp = &ptr[i];
}
printf("%s\n", tmp);
tmp = &blah[i]
可以与 互换tmp = &(*(blah + i))
。
您可以直接执行以下代码:
char *alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char *l = alpha;
如果您的代码如下:
const char *alpha = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const char *l = alpha;
:)
cpy 函数将接受两个 char 指针,src 指针将指向 main 函数中定义的 src(char array) 的初始字符,与 des 指针相同,将指向在 main 函数中定义的 des(char array) 的初始位置main 函数和 src 指针的 while 循环值会将值分配给 des 指针并将指针递增到下一个元素,这将发生直到 while 循环遇到 null 并退出循环,des 指针将在之后简单地分配 null取所有值。
#include<stdio.h>
void cpy(char *src,char *des)
{
while(*(des++) = *(src++));
*des = '\0';
}
int main()
{
char src[100];
char des[100];
gets(src);
cpy(src,des);
printf("%s",des);
}
输出:图像
#include <stdio.h>
#include <string.h>
#define MAX_LENGTH 256
int main(void) {
char *original, *copy, *start; //three character pointers
original = malloc(sizeof(char) * MAX_LENGTH); //assigning memory for strings is good practice
gets(original); //get original string from input
copy = malloc(sizeof(char) * (strlen(original)+1)); //+1 for \0
start = copy;
while((*original)!='\0')
*copy++ = *original++;
*copy = '\0';
copy = start;
printf("The copy of input string is \"%s\".",copy);
return 0;
}