I have a *char pointer and a char array. How can I put the value of the pointer in the char array? I've tried
uint8_t i;
char a[10];
char *b = "abcdefghi";
for(i = 0; i < 9; i++)
{
a[i] = b[i];
}
printf("%s", a);
But this doesn't work.
3 Answers
The size of a is 10, so is the length of b(including the null terminator). Change
for(i = 0; i < 9; i++)
to
for(i = 0; i < 10; i++)
6 Comments
philippe lhardy
yes strings are NULL terminated in C, last char be '0'.
Jite
@philippelhardy Not
'0', but '\0' or 0philippe lhardy
yes i should not have used enclosing '. 0x0 the byte 0.
Yu Hao
@Rio
i < 10 is more clear than i <= 9 in C-based languages, so it's preferred. |
What you are doing is right.
Juts change i<10 in your for loop.
There are 10 elements in your char array and the array index is from 0-9.
1 Comment
Ivan Ivanovich
strlen() in each iteration is bad ideayet another variant:
#include <stdio.h>
int main(void)
{
char a[10];
char *c = a;
char *b = "abcdefghi";
for(; *c = *b; ++c, ++b);
printf("%s", a);
return 0;
}
but strcpy() will work faster in any case, since it is often written in assembly language for a specific architecture
sprintf(a, "%p", b);to get the pointer into the buffera. However, it could be that the buffer is not wide enough, depending on the size of the pointer. Actually, check outsnprintf(), which should be safer.for(i = 0; a[i] = b[i]; i++);