Concatenating an int to a string or converting in C

问题

Im looking for a 6 digit random number on the end of foo-, I have been trying for a few hours now with no success. only error messages

note: expected 'char *' but argument is of type 'int'

Ive tried to convert the int to char but it just doesn't like it, My code is below,

#include <stdio.h>
#include <stdlib.h>    
#include <string.h>
#include <time.h>

char* concat(char *s1, char *s2)
{
    char *result = malloc(strlen(s1)+strlen(s2)+1);//+1 for the zero-terminator
    //in real code you would check for errors in malloc here
    strcpy(result, s1);
    strcat(result, s2);
    return result;
}

int main ()
{
  srand(time(NULL));
  int r = rand();

  printf(concat("foo-", r));

  return 0;
}

回答1:

for example:

srand(time(NULL));
int r = rand();
char* str = (char*)malloc(sizeof(char)*20);
snprintf(str, 7, "%d", r);

if (strlen(str) < 6) {  /* if r had less than 6 digits */
    sprintf(str, "%06d", r);
}

char* s = concat("foo-", str);
printf("%s",s);
free(str);
free(s);

return 0;

from http://joequery.me/code/snprintf-c/ :

int snprintf(char *str, size_t size, const char *format, ...);

str is the buffer where printf output will be redirected to. size is the maximum number of bytes(characters) that will be written to the buffer, including the terminating null character that snprintf automatically places for you. The format and the optional ... arguments are just the string formats like "%d", myint as seen in printf.

so, in order to get a 6 digit number converted, you specify in snprintf() the size argument to 7 (you include a null character).

sprintf() function sends formatted output to a string pointed to by the first argument (in our case this is str). %06d format provides that to str at least 6 digits will be sent.

CONCLUSION

you want to convert a 6 digit number to a char array. if the number had at least 6 digits, with snprintf() you will get the first 6 digits of the number that was converted. if it has less than 6 digits, the sprintf() will add zeroes at the beginning until there are 6 digits. so, if r = 101; the result would be 000101.

---

note that with your usage of concat function:

printf(concat("foo-",r));

compiler warns about (gcc in my case):

warning: format not a string literal and no format arguments [-Wformat-security]

printf() function needs to know the format of the string argument that is passed. there's a nice answer about it right here : warning: format not a string literal and no format arguments

回答2:

You could do this

printf("foo-%d", r);

or

char buffer[10];
sprintf(buffer, "%d", r);
char *c = concat("foo-", r);
printf("%s", c);
free(c);

This will use your function. Please read the manual pages for sprintf and printf

回答3:

In the example I'm using sprintf to convert int to char[]. Also I'm using modulo operation to ensure the result will have no more than 6 digits. The %06d argument for sprintf will ensure that the result will have no less than 6 digits.

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

int main() {
    char buf[7] = { 0 };
    int r;

    srand(time(NULL));

    r = rand() % 1000000;
    sprintf(buf, "%06d", r);

    printf("foo-%s\n", buf);

    return 0;
}

回答4:

Your problem occurs when you attempt to call concat with the int value r as the second argument. Notice your definition of concat:

char* concat(char *s1, char *s2);

The second argument is char *s2. In your program you attempt to call it with:

int r = rand();

printf(concat("foo-", r));

r is an integer. Try this to correct the problem:

int main ()
{
    srand(time(NULL));
    int r = rand();
    char str[] = "bar";

    printf(concat("foo-", str));

    return 0;
}

来源：https://stackoverflow.com/questions/24464745/concatenating-an-int-to-a-string-or-converting-in-c