Hello ,,,

I have an array of integers and i need to convert it to string ...

PlZ Help me as soon as Possible ...

## All 11 Replies

Could we see an example of the code you tried?

take a look here

you will need to loop through the array with a for loop

i have do a dynamic array of integers and i store the values.

``````for(int i=Count-1 ; i>=0 ; i--)
{
Digit[i]=(str[i]-'0');
}
``````

this is my code first i convert string to array then i need to convert it back to string after doing some operations >>>

plz help

Hello charming--eyes,
Do you want a string like "7456" if the integer array a[5] has a[0]=7,a[1]=4,a[2]=5,a[4]=6 ?

Yes>>

``````std::stringstream out;
for(size_t i=0; i < size; i++)
{
int x = digit[i];
x = x >> i; // do something to digit
out << x;
}``````

Hello charming--eyes,
The ASCII value of numeric characters start from 48 which refers to 0.Add the integer and 48,then store it in the character array.

``````int a[100];
char a[100];
a[0]=48+f[0];``````

For example if f[0] has 7,then 48 +7 = 55.So a[0] will take up the ASCII value as 55.
Put this in a loop. You will have the integers converted to string.

commented: Thanks +1

Thank you very Much >>Arbus>>

my problem has been solver

Err, I'm not the OP... I was trying to help him by pointing him to a method of combining strings.

``````**Basic Code in C Language**

#include<stdio.h>
#include<malloc.h>
void int_to_char_array(int *arr,int arrLen,char *str)
{
int count=0;
while(count<arrLen)
{
if(*(arr+count)/10!=0)
{
*str++=*(arr+count)/10+'0';
*(arr+count)%=10;
}
else
{
*str++=*(arr+count)%10+'0';
count++;
}
}
*str='\0';
}

void main()
{
int *arr,arrLen,count;
char *str;
puts("Enter size of int array :");
scanf("%d",&arrLen);
arr=(int *)malloc(sizeof(int)*arrLen);
str=(char *)malloc(sizeof(char)*arrLen);
puts("Enter elements of int array :");
for(count=0;count<arrLen;count++)
scanf("%d",(arr+count));
int_to_char_array(arr,arrLen,str);
puts("Main Result :");
puts(str);
}
``````
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