[英]Why can't I convert 'char**' to a 'const char* const*' in C?
[英]Can't assign const char* to char*
我正在用C编写并且必须返回一个char *,我正在尝试复制strcpy函数。 我有以下代码
int main()
{
char tmp[100];
char* cpyString;
const char* cPtr = &tmp[0];
printf("Enter word:");
fflush(stdin);
scanf("%s", &tmp);
cpyString = strcpy("Sample", cPtr);
printf("new count is %d\n", strlen(cpyString));
}
int strlen(char* s)
{
int count = 0;
while(*(s) != 0x00)
{
count++;
s = s+0x01;
}
return count;
}
char* strcpy(char* dest, const char* src)
{
char* retPtr = dest;
int i =0;
int srcLength = strlen(src);
for(i = 0; i< srcLength; i++)
{
*(dest) = *(src); //at this line program breaks
dest = dest + 0x01;
src = src + 0x01;
}
*(dest) = 0x00; //finish with terminating null byte
return retPtr;
}
Q1:如何在程序不崩溃的情况下将src处的解引用值分配给目标?
Q2:如果我需要将输入的tmp
字符串复制到新字符串中,该怎么办? 我似乎无法通过tmp
作为第二个参数
这里
cpyString = strcpy("Sample", cPtr);
^^^^^^^
const
您已交换参数。 第一个参数是不允许写入的字符串文字(“样本”)。 参见https://stackoverflow.com/a/4493156/4386427
尝试
cpyString = strcpy(cPtr, "Sample");
我不确定第二行是否正是您想要的,但至少是合法的。
也许您真的想要:
cpyStringBuffer[100];
cpyString = strcpy(cpyStringBuffer, cPtr);
通常,您的main
的代码比所需的更为复杂。
尝试:
int main()
{
char input[100] = {0};
char dest[100];
printf("Enter word:");
scanf("%99s", input); // notice the 99 to avoid buffer overflow
strcpy(dest, input);
printf("new count is %d\n", strlen(dest));
return 0;
}
我想您可能想像下面这样编码。
#include <stdio.h>
int strlen(char* s);
char* strcpy(char* dest, char* src);
int main()
{
char tmp[100];
char cpyString[100];
printf("Enter word:");
fflush(stdin);
scanf("%s", &tmp);
strcpy(cpyString, tmp);
printf("new count is %d\n", strlen(cpyString));
}
int strlen(char* s)
{
int count = 0;
while(*(s) != 0x00)
{
count++;
s = s+0x01;
}
return count;
}
char* strcpy(char* dest, char* src)
{
char* retPtr = dest;
int i =0;
int srcLength = strlen(src);
for(i = 0; i< srcLength; i++)
{
*(dest) = *(src); //at this line program breaks
dest = dest + 0x01;
src = src + 0x01;
}
*(dest) = 0x00; //finish with terminating null byte
return retPtr;
}
我认为您使用了未初始化的目标和文字字符串指针。 您必须将目的地声明为缓冲区,例如
char dest[const_size]
所以
char* strcpy(char* dest, const char* src)
{
char* retPtr = dest;
int i =0;
int srcLength = strlen(src);
for(i = 0; i< srcLength; i++)
{
*(dest) = *(src); //at this line program breaks
dest = dest + 0x01;
src = src + 0x01;
}
*(dest) = 0x00; //finish with terminating null byte
return retPtr;
}
int main()
{
char *arr="xxxxxx";
char *dest="fffff"; // this won't work because you can not modify const string
char *dest_1; // this won't work because it is uninitialized pointer
char dest_2[50]; // this will work fine
strcpy(x, y);
printf("%s",x);
//x still the same as point pointer
return 0;
}
您的程序崩溃,因为您无法将指针修改为常量。 请在下面找到更正的代码:
char *
mstrcpy (char *dest, const char *src)
{
char *retPtr = dest;
int i = 0;
int srcLength = strlen (src);
for (i = 0; i < srcLength; i++)
{
*(dest) = *(src); //now doesn't break at this line
dest = dest + 1;
src = src + 1;
}
*(dest) = 0x00; //finish with terminating null byte
return retPtr;
}
int
main ()
{
//char a = "abc"; // will cause crash
char a[] = "abc"; // won't crash
char *b = "xyz";
mstrcpy(a,b); //works fine !!!!
return 0;
}
请注意,在主函数中,如果使用char a = "abc"
,则会引起问题,因为其指向常量的指针
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.