How to check all strings?

[bitcoin]

There is a task, you need to compare a string with an array of strings. I don't know if I'm doing the right thing. It works, but I would like it to be somehow more elegant .. I think this solution is redundant. Or is it ok?

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char *data[] = { "data1", "data2", "data3", "data4", NULL };
char *p;
UINT i = 0;
do
{
p = data[i];
/* doing something with string */
i++;
} while (p!=NULL);

[CLR]

Hi bitcoin. I think it is OK.

I'd write

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for (int i = 0; data[i] != NULL; i++){ /* do something with data[i] */}


[30f7r8d2]

Excuse me please but why there is NULL in *data[], isn't it an array of strings so there should be no need to NULL terminate it?

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char *data[] = { "data1", "data2", "data3", "data4", NULL };

[TimoVJL]

This goes through strings.

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#include <stdio.h>

char *data[] = { "data1", "data2", "data3", "data4", 0 };

int __cdecl main(void)
{
int i = 0;
while (data[i]) {
int j = 0;
while (data[i][j]) {
printf("%c ", data[i][j]);
j++;
}
printf("\n");
i++;
}
return 0;
}

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d a t a 1
d a t a 2
d a t a 3
d a t a 4


[bitcoin]

TimoVJL , CLR
Thank you !

Excuse me please but why there is NULL in *data[], isn't it an array of strings so there should be no need to NULL terminate it?

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char *data[] = { "data1", "data2", "data3", "data4", NULL };

Because how I can guess , where is the end of array?
There may be a different number of string. Do not manually enter the count.

[AlexN]

You can try this:

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char *data[] = {"data1", "data2", "data3", "data4"};
count = sizeof(data)/sizeof(data[0]);
or make a macro:

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#define COUNT(x) (sizeof(x)/sizeof(x[0]))

char *data[] = {"data1", "data2", "data3", "data4"};
count = COUNT(data);