Append Formatted String to String in C (not C++) Without Pointer Arithmetic?

Question

Hi all,

Wrote a very simple C function to illustrate what I would like to simplify:

void main(int argc, char *argv[]){
    char *me="Foo";
    char *you="Bar";
    char us[100];
    memset(us,100,0x00);

    sprintf(us,"You: %s\n",you);
    sprintf(us+strlen(us),"Me: %s\n",me);
    sprintf(us+strlen(us),"We are %s and %s!\n",me,you);
    printf(us);
}

It seems like there should be a standard library function to handle what I'm doing with sprintf and advancing the pointer, no? Has been years since I've done any amount of C...

Thanks, -aj

Answer 1

You are probably looking for strcat (or even better strncat).

Answer 2

sprintf returns the number of non-NUL characters written.

int len = 0;
len += sprintf(us+len, ...);
len += sprintf(us+len, ...);
...

Answer 3

char *me="Foo";
char *you="Bar";
char us[100];

char* out = us;
out += sprintf(out,"You: %s\n",you);
out += sprintf(out,"Me: %s\n",me);
out += sprintf(out,"We are %s and %s!\n",me,you);
printf("%s", us);

Answer 4

Hmm...couldn't you just use something like:

sprintf(us, "You: %s\nMe: %s\nWe are %s and %s!\n, you, me, me, you);

Answer 5

Never use a non-constant string as the first argument to printf (or second argument to fprintf. This is bad practice. Try changing the me variable to "Muahaha %n%d%n%d%n%d%n%d%n" and say goodbye to your stack and say hello to SEGFAULT. printf will try and format the string but you haven't provided any arguments for it to deal with, and even though there are format specifiers in the string printf has no idea that you haven't given it any arguments.

If you want to output a 'pre-formatted' string, use fputs(str,stdout). There is rarely a case to use printf with a non-constant format string.