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a newcomer to c programming needs help again

i have once again come to a dead end with my logon program in c.
whatever value is given to my program it outputs the results for 2.
i have created a simpler program that produces the same error:

#include stdio

int main(int val1)
{
if(val1==1)
{
printf("1");
}
if(val1==2)
{
printf("2");
}
if(val1==3)
{
printf("3");
}
if(val1==4)
{
printf("4");
}
}

it is probably just me being stupid but can anyone explain this?

llewelm

The main() function takes two parameters as follows:

int main(int argc, char *argv[])

The first parameter is the number of command line arguments, the second are the actual arguments. My guess is that you've compiled your program and have a c.exe or something similar and your typing

c 1

and "2" is coming out. That is correct because your program has two command line arguments -- "c" and "1".

Your program should look more like:

#include <stdio.h>

int main(int argc, char *argv[])
{
if (strcmp(argv[1], "1") == 0) {
printf("1");
}
if (strcmp(argv[1], "2") == 0) {
printf("2");
}

etc...

ASKER CERTIFIED SOLUTION

PerryDK

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PerryDK

Heres the program written strictly in C code if you need it.

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char* argv[])
{
if(argc < 2)
{
printf("improper usage. Please pass some integer values to this program");

//return -1 indicating failure
return -1;
}

int numValues = argc -1;
int* ptrValues = (int*)malloc(numValues * sizeof(int));
for(int i = 0; i < numValues; i++)
{
//convert the paramters passed in to integers and store
//them in our array for later usage
//Note that if the string can not be converted to an integer
//the value stored will be 0
ptrValues[i] = atoi(argv[i+1]);
}

//lets enumerate through all the values displaying them to the user
//and lets add them up just for fun
int sum = 0;
for(int i = 0; i < numValues; i++)
{
printf("ptrValues[%i] = %i \n", i, ptrValues[i]);
sum += ptrValues[i];
}

printf("\nThe Sum of the numbers is %i", sum);

free(ptrValues);

return 0;
}