I am learning C. In the following code why does replacing "*ptr_str" with "ptr_str[i]" in the for loop truncate?
/* 13L01.c: Initiali开发者_开发问答zing strings */
#include <stdio.h>
main()
{
char str1[] = {'A', ' ',
's', 't', 'r', 'i', 'n', 'g', ' ',
'c', 'o', 'n', 's', 't', 'a', 'n', 't', '\0'};
char str2[] = "Another string constant";
char *ptr_str;
int i;
/* print out str2 */
for (i=0; str1[i]; i++)
printf("%c", str1[i]);
printf("\n");
/* print out str2 */
for (i=0; str2[i]; i++)
printf("%c", str2[i]);
printf("\n");
/* assign a string to a pointer */
ptr_str = "Assign a strings to a pointer.";
for (i=0; *ptr_str; i++)
printf("%c", *ptr_str++);
return 0;
}
// A, ok
while (*ptr_str)
printf("%c", *ptr_str++);
// B, also ok
for (i=0; ptr_str[i]; i++)
printf("%c", ptr_str[i]);
// C, works but ugly
for (i=0; *ptr_str; i++)
printf("%c", *ptr_str++);
C is your form, it is flawed because i is doing nothing here, so A is an improved version. If you want to use i, do it like B. If you use both i and ptr_str, and increment both of them in the loop, nothing good will happen. Increment one or the other.
Because you're advancing ptr_str
and then treating it as an array and testing if it's pointing to NULL
on the i'th member. You're basically testing if ptr_str[i+i]
is NULL
instead of ptr_str[i]
.
Your question is a bit unclear, but I think I've got it:
In the following code why does replacing "*ptr_str" with "ptr_str[i]" in the for loop truncate?
I think you mean changing this:
for (i=0; *ptr_str; i++)
printf("%c", *ptr_str++);
return 0;
to this:
for (i=0; ptr_str[i]; i++)
printf("%c", *ptr_str++);
return 0;
The second one truncates because you're advancing i
and ptr_str
, so the modified starting position of ptr_str
plus the modified starting position of i
ends up cutting you off too soon (or worse, having an odd number of characters and overflowing into data that isn't yours). The second example that truncates is equivalent to:
for (i=0; ptr_str[i * 2]; i++)
printf("%c", ptr_str[i]);
return 0;
Now do you see why it truncates?
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