Hi..
Heres an example of a code:
now i want to copy info to p1 so I would get string with "is some text and another text..";Code:char data[128] = "here is some text"; char info[128] = " and another text.."; char *p1, p2; p1 = data; p1 += 5;
What would be the proper way to do that?
Thanks a lot for your help guys
the =+ works with strings, ie:
string One = "This is a ";
string Two = "Complete sentence!!";
string Three = One + Two;
One += Two;
etc
Use strcpy and strcat -- but you'll need to allocate some memory to p1 to be able to write to it.
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
So I could do:
and i would get printf(p1) : "here is some text and another text.." ?Code:char data[128] = "here is some text"; char info[128] = " and another text.."; char *p1, p2; p1 = data; p2 = info; p1 += p2;
What way would you prefer:
orCode:sprintf(tmp, "%s%s", p1, p2);
Whats more professional?Code:strcpy(tmp, p1); strcat(tmp, p2);
Last edited by l2u; 05-19-2006 at 09:00 AM.
>>What would be the proper way to do that?<<
Use std::string, not char arrays as twomers has already described.
>>printf<<
No, that would be std::cout. This is the c++ board, char arrays and printf are for c programming.
Code:#include <iostream> #include <string> int main() { std::string data = "here is some text"; std::string info = " and another text.."; data+=info; std::cout<<data<<std::endl; }
Last edited by Ken Fitlike; 05-19-2006 at 12:11 PM. Reason: typo
CProgramming FAQ
Caution: this person may be a carrier of the misinformation virus.
If you wanted to do it in C, you could
I don't know what you wanted to do with p2, but if you wanted to define two pointers, you should haveCode:#include <stdio.h> #include <string.h> #include <malloc.h> int main() { char data[128] = "here is some text"; char info[128] = " and another text.."; char *p1, p2; p1 = (char*) malloc((strlen(data) + strlen(info) + 1) * sizeof(char)); strcpy(p1, data); strcat(p1, info); printf("%s\n", p1); free(p1); return 0; }
Your code would have defined one character pointer and one character.Code:char *p1, *p2;
If you're going to do it in C, there is a perfectly good function for something like this.
Code:#include <stdio.h> #include <stdlib.h> #include <string.h> int main(void) { char data[] ="Here is some text "; char info[] = "which is a part of a bigger sentence."; /* I cached the length of both strings to reduce some overhead */ const int DATALEN = strlen(data); const int INFOLEN = strlen(info); char *sentence = malloc(DATALEN + INFOLEN + 1); sprintf(sentence, "%*s%*s", DATALEN, data, INFOLEN, info); puts(sentence); free(sentence); return 0; }
Last edited by whiteflags; 05-19-2006 at 11:11 AM.
Why is * sizeof(char) needed?
It's not. It is the same as multiplying by one.Originally Posted by l2u
7. It is easier to write an incorrect program than understand a correct one.
40. There are two ways to write error-free programs; only the third one works.*
