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[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編寫並且必須返回一個char *,我正在嘗試復制strcpy函數。 我有以下代碼
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:如何在程序不崩潰的情況下將src處的解引用值分配給目標?
Q2:如果我需要將輸入的tmp
字符串復制到新字符串中,該怎么辦? 我似乎無法通過tmp
作為第二個參數
這里
cpyString = strcpy("Sample", cPtr);
^^^^^^^
const
您已交換參數。 第一個參數是不允許寫入的字符串文字(“樣本”)。 參見https://stackoverflow.com/a/4493156/4386427
嘗試
cpyString = strcpy(cPtr, "Sample");
我不確定第二行是否正是您想要的,但至少是合法的。
也許您真的想要:
cpyStringBuffer[100];
cpyString = strcpy(cpyStringBuffer, cPtr);
通常,您的main
的代碼比所需的更為復雜。
嘗試:
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這樣編碼。
#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;
}
我認為您使用了未初始化的目標和文字字符串指針。 您必須將目的地聲明為緩沖區,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;
}
您的程序崩潰,因為您無法將指針修改為常量。 請在下面找到更正的代碼:
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;
}
請注意,在主函數中,如果使用char a = "abc"
,則會引起問題,因為其指向常量的指針
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