Input Loop trouble [closed] - c

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What I've done is ask a user to input how many times the program will loop, it then records values into 3 different arrays. Everything is working great, but what I need it to do is print the elements of one array if the corresponding element in another array meets the requirements. Everything else runs great, I'll post the two arrays that I'm trying to use for this.
char *names[50][32];
char *states[50][2];
i = 0;
while ( i < b) {
if (state[i] = "tx");{
printf("a string %s\n", names[i]);}
i = i + 1;
}
for this : if (state[i] = "tx");{ I've tried with and without quotes and using 116120...
Basically, it asks for peoples names and where they live. I can get it to print the array element values for each name(it runs in a loop) but I want it to only print the names for the people who live in tx.

Answer 1

There are a few things wrong with your code. First of all ending an if or for construction with a semicolon is a common mistake when starting in C. Basically it creates an empty if statement followed by a code block. Look at it this way:
if (condition)
; // Does nothing. The if is empty
// Totally unrelated block of code.
{
}
Code blocks are usually useful to create scopes, so although it might seem useless for the compiler to interpret blocks in this way, it actually is not. This also happens in other situations, such as while, for, and so on:
for (int i=0 ; i<n ; ++i)
; // Empty for. Runs `n` loops, but doing nothing
// Unrelated block of code. Runs only once
{
}
The comparison operator is also wrong, you should use == for comparisons, instead of =, which is used for assignments.
Finally, you cannot compare strings in this way. Strings are basically arrays, which in turn are represented using pointers. If you compare two pointers (ptr1 == ptr2) it'll only check whether the two strings point at the same address in memory. As strings are composed by several characters, they have to be iterated to be properly compared. Fortunately the standard library already provides a method for this.
Fix a typo or two and this is what you get:
char *names[50][32];
char *states[50][2];
i = 0; // Assuming this is declared somewhere else
while ( i < b) {
if (strcmp(states[i], "tx") == 0) {
printf("a string %s\n", names[i]);
}
i = i + 1;
}
You should probably also check the docs for strcmp.
Edit: as this is already the accepted answer, I should also include a fix as noted by #dbush. The array for stats is clearly missing space for the extra string terminator, as strings are NULL terminated in C. The array for names might or might not suffer from the same issue, it's not clear. Anyway, it's notable that both should include an extra byte for storing the terminator:
char names[50][33];
char states[50][3];
Props to #dbush.

Answer 2

The declaration of your arrays don't look correct:
char *names[50][32];
char *states[50][2];
These declare a pair of two-dimensional array of char pointers, which is probably not what you want.
char names[50][32];
char states[50][3];
These are two-dimensional arrays of characters, or alternately arrays of strings. Note that the states array has space for an extra character for the terminating NULL.
In this if statement this:
if (state[i] = "tx");{
Since the ; occurs immediately after the condition, that ends the if block. The following statements within curly braces therefore will always run. Also, = is for assignment, not comparison, but using == is not appropriate either, since that operator won't compare the strings, but their addresses. You need to use strcmp for string comparisons.
So the fixed code should look like this:
char names[50][32];
char states[50][3];
...
i = 0;
while ( i < b) {
if (strcmp(state[i],"tx") == 0) {
printf("a string %s\n", names[i]);
}
i = i + 1;
}