16

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);
}
4

7 回答 7

19

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);
于 2013-09-21T23:40:26.667 回答
17

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;
}
于 2013-09-22T00:28:10.123 回答
3

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lwith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/

于 2013-09-21T23:45:49.160 回答
0

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))

于 2015-06-29T16:32:34.060 回答
0

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;

:)

于 2016-11-15T10:30:50.203 回答
0

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

于 2017-08-11T13:23:12.263 回答
0
#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;
}
于 2015-09-17T05:48:54.427 回答