Create a string from a vector of integers in C - c

Question

I'm trying to create a string from a vector, something like this:
int vector[3] = {1, 2, 3}.
Then, the string I would get should be:
[1, 2, 3] stored in some char[] variable.
I tried to do it dynamically like follows:
void print_vector(int *vector, int lon){
char *str = (char*)calloc(3*lon, sizeof(char));
str[0] = "[";
str[1] = (char) vector[0];
for(int i = 1; i < lon; i += 3){
str[i] = (char)vector[i];
if(i != lon - 1){
str[i + 1] = (char)",";
str[i + 2] = " ";
}else{
str[i + 1] = "]";
}
}
printf("%s\n", str);
}
Where lon is the length of the vector, but I'm getting this warning from the compiler:
warning: the assignment to "char" from "char *" creates an integer from a pointer without a conversion [-Wint-conversion]
What am I doing wrong? Thanks.

Answer 1

There are several issues with your code -
You are using " instead of '. " is used to create a string literal which is of type char*. Hence the error you are getting.
(char) vector[0]; is not the right way to convert an integer to strings. You should use something like sprintf.
You haven't allocated enough memory with the call to calloc (the brackets, commas, space, nul-terminator).
You are only filling in certain characters in the buffer. The rest are initialized to 0. This will cause the string to terminate earlier than expected.
To start, you can fix your code as -
void print_vector(int *vector, int lon){
char *str = calloc(5*lon+3, sizeof(char));
// 5 for each character (3 characters per integer + space + comma.
// Two brackets and nul-terminator
strcpy(str, "[");
if (lon) {
sprintf(str+1, "%d", vector[0]);
for(int i = 1; i < lon; i++){
sprintf(str + strlen(str), ", %d", vector[i]);
}
}
strcat(str, "]");
printf("%s\n", str);
}
You can find the working demo here on ideone.

Answer 2

You're trying to assign strings to char array elements. You need to use single quotes to create char constants.
So
str[0] = "[";
should be
str[0] = '[';
and similarly for all the other assignments.
Another problem is that (char)vector[i] will not return the character representation of the number in vector[i]. It simply uses the integer as the character code. So if vector[i] is 1, this produces '\1', not '1'. The way to get the character representation of a number is with '0' + vector[i].
Finally, you need to add a null terminator to the end of the string so you can print it with %s formatting. And you should free it before returning to prevent a memory leak.
Other problems: You're only putting every 3rd element of vector into the string. You need to use separate indexes for vector and str. And array indexes start at 0, not 1.
void print_vector(int *vector, int lon){
char *str = (char*)calloc(3*lon, sizeof(char));
str[0] = "[";
str[1] = (char) vector[0];
for(int i = 0; i < lon; i++){
str[i*3] = '0' + vector[i];
if(i != lon - 1){
str[i*3 + 1] = ',';
str[i*3 + 2] = ' ';
}else{
str[i*3 + 1] = ']';
str[i*3 + 2] = '\0';
}
}
printf("%s\n", str);
}

Answer 3

Assuming every element in your vector is one digit, and every element is spaced identicially...
... and remembering you need to allicate a null byte at the end to terminate your string...
... perhaps something like this (uncompiled/untested):
void print_vector(int *vector, int lon){
char *str = (char*)calloc((3*lon)+1, sizeof(char));
char *s = str;
*s++ = "[";
int i=1;
while (i < lon) {
*s++ = vector[i] + 0x30; // ASCII "0"= 0x30, "1"= 0x31, ...
*s++ = ',';
*s++ = ' ';
i++;
}
s -= 2;
*s++ = ']';
*s++ = 0;
printf("%s\n", str);