char array[12];
sprintf(array, "%s %s", "Hello", "World");
printf(array); // prints "Hello World"
Is there a way to do it without using sprintf or strcpy?
5 Answers
You cannot assign a string to an array directly, but you can initialize an array from a string literal:
char array[] = "Hello World"; // this defines array with a size of 12 bytes
If you later want to store a different string to array, you must use a string copy function such as these:
strcpy(array, "Hello Buddy"); // assuming array has at least 12 bytes
memcpy(array, "Hello Buddy", 12); // assuming array has at least 12 bytes
snprintf(array, sizeof array, "%s %s", "Hello", "Buddy");
or you can assign characters one at a time:
array[6] = 'B';
array[7] = 'u';
array[8] = 'd';
array[9] = 'd';
array[10] = 'y';
array[11] = '\0';
Note that it is preferred to use snprintf instead of sprintf to avoid potential buffer overruns. Also avoid passing a variable array to printf, which would cause undefined behavior if it contains % characters. Always use a constant format string:
printf("%s\n", array);
2 Comments
Mad Physicist
How about a loop?
Mnkisd
So, basically, it's not possible in a handy way (without strcpy and snprintf) , since assigning EACH char value is not an handy way. Thanks for reply!
You can do
char array[] = "Hello World";
1 Comment
Jabberwocky
That's not an assignment, it's an initialization
Given two strings and a buffer big enough to hold both, you can use a loop to copy the content:
const char *s1 = "Hello";
const char *s2 = "World";
char s[12];
int i;
for(i = 0; i < 5; i++) {
s[i] = s1[i];
}
s[i++] = ' ';
for(i = 0; i < 5; i++) {
s[i + 6] = s2[i];
}
s[i + 6] = '\0';
If it's a case like command line arguments, where you have an array of pointers with a count, you can use a pair of nested loops for the whole thing:
char *argv[] = {
"Hello",
"World",
NULL
};
int argc = 2;
char s[12];
int i, j, k;
j = 0;
for(k = 0; k < argc; k++) {
for(i = 0; argv[k][i]; i++) {
s[j++] = argv[k][i];
}
s[j++] = ' ';
}
if(k) {
s[--j] = '\0';
} else {
s[0] = '\0';
}
1 Comment
Mad Physicist
@chqrlie. Thanks for those catches. I believe I've fixed them. I appreciate the debugging at a time when I couldn't.
a way to assign a string to an array:
#define MAX_LEN 12
int main( void )
{
...
char array[ MAX_LEN ];
...
memset( array, '\0', MAX_LEN );
memcpy( array, "hello world", strlen( "hello world" ) );
5 Comments
Jabberwocky
This is just a convoluted and totally unsafe way do avoid strcpy, so it's pretty pointless. And the memset could be avoided, providing you make a slight changement in the line with memcpy.
user3629249
@Jabberwocky, The OP ask how to do it without using
strcpy() That is exactly what I provided, So this answer exactly answers the OPs question. So it is not pointless. I never indicated it was a 'safe' methodJabberwocky
We should not give poor advice here.
user3629249
@Jabberwocky, I could have declared the array via:
char array[ MAX_LEN ] = {\0'}; But decided not to do so for flexibility/reusability.Jabberwocky
What I mean to say is not using
strcpy is a poor choice. His question is like, I want to add two numbers but I don't want to use the + operator. I can be done, but why?Eventhough I could not understand what you really meant, from your title, I understand that you ask how to assign a value to a string. Thus I will write the code of it. I hope it helps.
char array[] = "Hello World!";
or you can assign more than one values like:
char array[12];
array[0]="H";
array[1]="e";
and can go till the last point. Yet they will be static, you will not be able to copy them by using strcpy. If you want, you have to use malloc which is memory allocation for the array of strings.
1 Comment
mcleod_ideafix
You are assigning pointers to strings to elements 0 and 1 of array. Well, sort of, because they most probably won't fit in them.
array[0]='H'; array[1]='e'; /*...*/ array[11]=0;strcpy?printf(array);is bad practice, btw. What ifarrayhas a percent sign in it? Usefputs(array, stdout);orprintf("%s", array);instead to print a string to standard output.