I wrote a string copy routine for learning purposes and then I dellibratly made it leave the \0 off the destination string and it still works why is this?
Code:char *chrstrcpy(char *des,char *src) { char *p = des; while (*src != '\0') { *p++ = *src++; } return des; }
it does not work for me. maybe you forgot to recompile your program
You need to define what you mean by works and does work.
Code:#include <stdio.h> char *chrstrcpy(char *des, char *src) { char *p = des; while (*src != '\0') { *p++ = *src++; } return des; } int main(void) { char s[10] = "aaaaaaaa"; puts(s); chrstrcpy(s, "BBBB"); puts(s); return 0; } /* Output aaaaaaaa BBBBaaaa */
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would the \0 clear the end of the old string?
Try it and find out.Originally posted by Gav D
would the \0 clear the end of the old string?
When all else fails, read the instructions.
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sorry I know it then terminate the string and the program wouldnt show the left over value form the old variable but would it still be in memory or would that be cleared?
>>would it still be in memory or would that be cleared?
Again, try it and find out. You'll need something like this:
Short answer though, nothing is "cleared" unless you specifically write over it.Code:for (i = 0; i < arraysize; i++) { putchar (array[i]); }
When all else fails, read the instructions.
If you're posting code, use code tags: [code] /* insert code here */ [/code]
