For example, I have a string containing "184553", how do I go about extracting the 3rd character "4" from string string and assigning it to another variable?
I have tried the following code but it does not work, the system crashes:
char string[6] = "184553";
printf("%s",string[2]); // Expecting it to print "4"Any help??
Like so
char my_ch = string[2];
hi there,
As you want to print a single character please use %c as the format specifier instead of %s. Also the size of string variable should be 7.
The code becomes as:
char string[7] = "184553";
printf("%c",string[2]); //prints "4" nowAnirudh
Yes, %c should be used instead of %s. This isn't technically a C-string, just a simple array. A string would require 7 spaces, to accommodate for the NULL value at the end.
Yes, %c should be used instead of %s. This isn't technically a C-string, just a simple array. A string would require 7 spaces, to accommodate for the NULL value at the end.
No, technically it is a C-string. It has quotes which by definition ends in a NULL value. The definition actually overflows the array.
My mistake
No, technically it is a C-string. It has quotes which by definition ends in a NULL value. The definition actually overflows the array.
Thank-you WaltP
I appreciate your technical prowess with these questions..and I hope you realize this is sincere...
When you want to declare an array of the same size as a const string literal then don't specify the array size so that the compiler can figure out the size itself char string[] = "184553";
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