I am having an array of integer say int example[5] = {1,2,3,4,5}. Now I want to convert them into character array using C, not C++. How can I do it?
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2What does 1,2,3,4,5 stand for? What are some reasonable conversions you are expecting? Please add an example to your question.Aamir– Aamir2011-01-17 11:46:11 +00:00Commented Jan 17, 2011 at 11:46
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Its an typo... its int example[5]CrazyCoder– CrazyCoder2011-01-17 11:48:40 +00:00Commented Jan 17, 2011 at 11:48
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@winbros - char array in the sense you want it as {'1','2','3','4','5'} ?Sachin Shanbhag– Sachin Shanbhag2011-01-17 11:51:17 +00:00Commented Jan 17, 2011 at 11:51
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Do you want to deal with ASCII of 1,2...?Javed Akram– Javed Akram2011-01-17 11:55:54 +00:00Commented Jan 17, 2011 at 11:55
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5 Answers
Depending on what you really want, there are several possible answers to this question:
int example[5] = {1,2,3,4,5};
char output[5];
int i;
Straight copy giving ASCII control characters 1 - 5
for (i = 0 ; i < 5 ; ++i)
{
output[i] = example[i];
}
characters '1' - '5'
for (i = 0 ; i < 5 ; ++i)
{
output[i] = example[i] + '0';
}
strings representing 1 - 5.
char stringBuffer[20]; // Needs to be more than big enough to hold all the digits of an int
char* outputStrings[5];
for (i = 0 ; i < 5 ; ++i)
{
snprintf(stringBuffer, 20, "%d", example[i]);
// check for overrun omitted
outputStrings[i] = strdup(stringBuffer);
}
3 Comments
Alston
What if the integer number is larger than
9? Integers above 10 won't be converted to the char format.
ilja
Is there a solution without a for loop for the first option (ASCII control characters), or is this already the cleanest solution?
JeremyP
@ilja I can't think of one.
#include <stdio.h>
int main(void)
{
int i_array[5] = { 65, 66, 67, 68, 69 };
char* c_array[5];
int i = 0;
for (i; i < 5; i++)
{
//c[i] = itoa(array[i]); /* Windows */
/* Linux */
// allocate a big enough char to store an int (which is 4bytes, depending on your platform)
char c[sizeof(int)];
// copy int to char
snprintf(c, sizeof(int), "%d", i_array[i]); //copy those 4bytes
// allocate enough space on char* array to store this result
c_array[i] = malloc(sizeof(c));
strcpy(c_array[i], c); // copy to the array of results
printf("c[%d] = %s\n", i, c_array[i]); //print it
}
// loop again and release memory: free(c_array[i])
return 0;
}
Outputs:
c[0] = 65
c[1] = 66
c[2] = 67
c[3] = 68
c[4] = 69
1 Comment
JeremyP
Your answer will only work for numbers up to 999, after that it'll start truncating digits - INT_MAX for a 32 bit 2's complement integer is 2147483647 which as rather more than 3 digits (one of your four is taken with
'\0')You can convert a single digit-integer into the corresponding character using this expression:
int intDigit = 3;
char charDigit = '0' + intDigit; /* Sets charDigit to the character '3'. */
Note that this is only valid, of course, for single digits. Extrapolating the above to work against arrays should be straight-forward.
Comments
You need to create the array, because sizeof(int) is (almost surely) different from sizeof(char)==1.
Have a loop in which you do char_example[i] = example[i].
If what you want is to convert an integer into a string you could just sum your integer to '0' but only if you're sure that your integer is between 0 and 9, otherwise you'll need to use some more sophisticated like sprintf.
Comments
In pure C I would do it like this:
char** makeStrArr(const int* vals, const int nelems)
{
char** strarr = (char**)malloc(sizeof(char*) * nelems);
int i;
char buf[128];
for (i = 0; i < nelems; i++)
{
strarr[i] = (char*)malloc(sprintf(buf, "%d", vals[i]) + 1);
strcpy(strarr[i], buf);
}
return strarr;
}
void freeStrArr(char** strarr, int nelems)
{
int i = 0;
for (i = 0; i < nelems; i++) {
free(strarr[i]);
}
free(strarr);
}
void iarrtostrarrinc()
{
int i_array[] = { 65, 66, 67, 68, 69 };
char** strarr = makeStrArr(i_array, 5);
int i;
for (i = 0; i < 5; i++) {
printf("%s\n", strarr[i]);
}
freeStrArr(strarr, 5);
}
1 Comment
user411313
A malloc-cast isn't 'pure C', its C++.
