abort trap: 6 error while running this program in C - c

Question

I am writing this code that takes a calculation from a file on the command line, then calculates it. I have gotten to a point where I am able to get each part of the file into a string array, however when I try to run that array into a method to calculate it, I get an abort trap: 6 error during run time. I am writing in c using emacs and the cc in the terminal to compile.
I am running this in main when it gives the error, with these variables initialized:
char items[10][10];
int sig;
int numOfItems = n-1;
double result[3];
sig=calculate(items, numOfItems, result);
The method is the following, I know it does not fully work yet, but it seems to not even enter the method because it does not print out the test print statements I have put:
int calculate(char items[10][10], int numOfItems, double *res){
printf("test2");
int flag, i, c, j, numdigits, decindicate, operatorindicate,n,m;
double ans;
double numbers[30];
char operators[30];
for(n=0; n<30; n++){
numbers[n]=0;
}
flag=1;
i=0;
numdigits=0;
decindicate=0;
operatorindicate=0;
n=0;
m=0;
j=0;
while(n<numOfItems){
c=items[n][0];
if(('0'<=c) && (c<='9')){
m=0;
while(c!='\0'){
c=items[n][m];
if(('0'<=c) && (c<='9')){
numbers[i] = numbers[i]*10+(c-'0');
}else if(c=='.'){
decindicate=1;
m++;
break;
}else{
flag=0;
break;
}while(decindicate>0){
c=items[n][m];
if(('0'<=c) && (c<='9')){
numbers[i]=numbers[i]+((c-'0')/(10^decindicate));
decindicate++;
}else{
flag=0;
break;
}m++;
}
i++;
}
}else if(c=='+' || c=='-' || c=='*' || c=='/'){
operators[j]=c;
j++;
}else{
flag=0;
break;
}n++;
}
for(i=0; i<numdigits; i++){
printf("%lf \n",numbers[i]);
}for(i=0; i<m; i++){
printf("char %c \n", operators[i]);
}
ans=numbers[0];
for(i=0; i<n-1; i++){
if(operators[i]=='+'){
ans=ans+numbers[i+1];
}else if(operators[i]=='-'){
ans=ans-numbers[i+1];
}else if(operators[i]=='*'){
ans=ans*numbers[i+1];
}else if(operators[i]=='/'){
ans=ans/numbers[i+1];
}
}
*res=ans;
return flag;
}
Here is the entire code if it's something out of these bounds that I don't even realize:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int calculate(char items[10][10], int numOfItems, double *res){
int flag, i, c, j, numdigits, decindicate, operatorindicate,n,m;
double ans;
double numbers[30];
char operators[30];
for(n=0; n<30; n++){
numbers[n]=0;
}
flag=1;
i=0;
numdigits=0;
decindicate=0;
operatorindicate=0;
n=0;
m=0;
j=0;
while(n<numOfItems){
c=items[n][0];
if(('0'<=c) && (c<='9')){
m=0;
while(c!='\0'){
c=items[n][m];
if(('0'<=c) && (c<='9')){
numbers[i] = numbers[i]*10+(c-'0');
}else if(c=='.'){
decindicate=1;
m++;
break;
}else{
flag=0;
break;
}while(decindicate>0){
c=items[n][m];
if(('0'<=c) && (c<='9')){
numbers[i]=numbers[i]+((c-'0')/(10^decindicate));
decindicate++;
}else{
flag=0;
break;
}m++;
}
i++;
}
}else if(c=='+' || c=='-' || c=='*' || c=='/'){
operators[j]=c;
j++;
}else{
flag=0;
break;
}n++;
}
for(i=0; i<numdigits; i++){
printf("%lf \n",numbers[i]);
}for(i=0; i<m; i++){
printf("char %c \n", operators[i]);
}
ans=numbers[0];
for(i=0; i<n-1; i++){
if(operators[i]=='+'){
ans=ans+numbers[i+1];
}else if(operators[i]=='-'){
ans=ans-numbers[i+1];
}else if(operators[i]=='*'){
ans=ans*numbers[i+1];
}else if(operators[i]=='/'){
ans=ans/numbers[i+1];
}
}
*res=ans;
return flag;
}
int main(int argc, char **argv){
int n, m, i;
char digits[50]; //used to store digits before adding necessary spaces
char items[10][10]; //will store string array of items in equation
FILE *fp, *fp2;
int sig; //will indicate if there is an invalid character
double result[3] = {0}; //will return result of equation
fp = fopen(argv[1], "r");
if(fp==NULL){
printf("Please provide file");
}
fp2 = fopen("temp", "w+"); //will read to file temp
n=0;
while(0==0){
digits[n]=fgetc(fp);
if(digits[n]==EOF){
digits[n]='\0';
break;
}
n++;
}
n=0;
char temp1;
char temp2;
while(digits[n]!='\0'){
if((('0'<=digits[n]) && (digits[n]<='9') && (digits[n+1]=='+' || digits[n+1]=='-' || digits[n+1]=='*' || digits[n+1]=='/')) || ((digits[n]=='+' || digits[n]=='-' || digits[n]=='*' || digits[n]=='/') && (('0'<=digits[n+1]) && (digits[n+1]<='9')))){
temp1=digits[n+1];
digits[n+1]=' ';
m=n+2;
while(digits[m-1]!='\0'){
temp2=temp1;
temp1=digits[m];
digits[m]=temp2;
m++;
}
}
fputc(digits[n], fp2);
n++;
}
//test if digit array fills correctly
n=0;
while(digits[n]!='\0'){
printf("testings digits array: %c \n", digits[n]);
n++;
}
//scans the temp file to form string array
rewind(fp2);
i=1;
n=0;
while(i==1){
i=fscanf(fp2, "%s", items[n]);
n++;
}
int numOfItems = n-1;
//test if char array items fills correctly
n=0;
while(n<numOfItems){
printf("testing items array: %s \n", items[n]);
n++;
}
sig=calculate(items, numOfItems, result);
if (sig==0){
printf("This is not a valid operation. \n");
}else {
printf("The calculation equals %lf \n", result[0]);
}
remove("temp");
}
the file I am using to test is sc1 which contains the following:
34 + 96 - 10 / 2
This is what the whole program prints out when using this sc1 file:
testings digits array: 3
testings digits array: 4
testings digits array:
testings digits array: +
testings digits array:
testings digits array: 9
testings digits array: 6
testings digits array:
testings digits array: -
testings digits array:
testings digits array: 1
testings digits array: 0
testings digits array:
testings digits array: /
testings digits array:
testings digits array: 2
testing items array: 34
testing items array: +
testing items array: 96
testing items array: -
testing items array: 10
testing items array: /
testing items array: 2
Abort trap: 6
I feel so lost, if someone could help that would be great.

Answer 1

in calculate the while(c!='\0') was an infinite loop. c=items[n][m]; never got a new value for c as m wasn't incremented unless a dot was detected. i was incremented so each iteration accessed numbers[i] and when i exceeded the boundary of numbers[30] it failed.
numbers[i]=numbers[i]+((c-'0')/(10^decindicate)); is another problem as ^ is bit XOR. Not what you want here to raise the power of 10.
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
int calculate(char items[10][10], int numOfItems, double *res){
int flag = 1, i = 0, c, ops = 0, decindicate = 0, nums = 0,item = 0,each = 0;
double ans;
double numbers[30] = { 0.0};
char operators[30];
printf ( "calculating\n");
while ( item < numOfItems) {
c = items[item][0];
if ( ( '0' <= c && c <= '9') || c == '.') {
each = 0;
decindicate = 0;
while ( '\0' != ( c = items[item][each])) {
if ( ( '0' <= c) && ( c <= '9')) {
numbers[nums] = numbers[nums] * 10 + ( c - '0');
if ( decindicate) {
decindicate++;
}
}
else if ( c == '.') {//found a dot
decindicate = 1;//set to 1
each++;
continue;
} else {
flag = 0;
break;
}
each++;//advance each character
}
while ( decindicate > 1) {
decindicate--;
numbers[nums] /= 10;//divide by power of 10 digits after dot
}
nums++;//advance nums
if ( nums >= 30) {
printf ( "too many numbers\n");
return 0;
}
} else if ( c == '+' || c == '-' || c == '*' || c == '/') {
operators[ops] = c;
ops++;//sdvance ops
if ( ops >= 30) {
printf ( "too many operators\n");
return 0;
}
} else {
flag = 0;
break;
}
item++;//advance item
}
for ( i = 0; i < nums; i++) {
printf ( "numbers %f \n", numbers[i]);
}
for ( i = 0; i < ops; i++) {
printf ( "operators %c \n", operators[i]);
}
ans = numbers[0];
for ( i = 0; i < item - 1; i++) {
switch ( operators[i]) {
case '+':
ans = ans + numbers[i + 1];
break;
case '-':
ans = ans - numbers[i + 1];
break;
case '*':
ans = ans * numbers[i + 1];
break;
case '/':
ans = ans / numbers[i + 1];
break;
}
}
*res = ans;
return flag;
}
int main ( int argc, char **argv) {
int n = 0, i, digit = 0;
char digits[50] = ""; //used to store digits before adding necessary spaces
char items[10][10] = { { ""}}; //will store string array of items in equation
FILE *fp = NULL, *fp2 = NULL;
int sig = 0; //will indicate if there is an invalid character
int space = 1;//skip leading whitespace
int operator = 1;//must have a number first
int number = 0;
int dot = 0;
double result = 0.0f; //will return result of equation
if ( NULL == ( fp = fopen ( argv[1], "r"))) {
printf ( "Please provide file\n");
return 0;
}
if ( NULL == ( fp2 = fopen ( "temp", "w+"))) { //will read to file temp
printf ( "could not open temp file\n");
fclose ( fp);
return 0;
}
n = 0;
while ( EOF != ( digit = fgetc ( fp))) {
digits[n] = digit;
n++;
if ( n >= 49) {
break;
}
}
digits[n] = '\0';
fclose ( fp);
n = 0;
while ( digits[n] != '\0') {
switch ( digits[n]) {
case ' ':
case '\t':
case '\n':
if ( space) {
break;
}
space = 1;
//do not reset so as to detect consecutive numbers or operators
//number = 0;
//operator = 0;
dot = 0;
fputc ( digits[n], fp2);
break;
case '.':
if ( dot) {
break;
}
if ( space && number) {
printf ( "bad format. expected operation\n");
return 0;
}
if ( !number) {
fputc ( ' ', fp2);
}
space = 0;
number = 1;
operator = 0;
dot = 1;
fputc ( digits[n], fp2);
break;
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9':
if ( space && number) {
printf ( "bad format. expected operation\n");
return 0;
}
if ( !number) {
fputc ( ' ', fp2);
}
space = 0;
number = 1;
operator = 0;
dot = 0;
fputc ( digits[n], fp2);
break;
case '+':
case '-':
case '*':
case '/':
if ( space && operator) {
printf ( "bad format. expected number\n");
return 0;
}
if ( !operator) {
fputc ( ' ', fp2);
}
number = 0;
operator = 1;
dot = 0;
fputc ( digits[n], fp2);
space = 1;
fputc ( ' ', fp2);
break;
}
n++;
}
//scans the temp file to form string array
rewind(fp2);
i = 1;
n = 0;
while ( i == 1) {
i = fscanf ( fp2, "%s", items[n]);
n++;
if ( n >= 10) {
break;
}
}
fclose ( fp2);
int numOfItems = n-1;
//test if char array items fills correctly
n = 0;
while ( n < numOfItems) {
printf ( "testing items array: %s \n", items[n]);
n++;
}
sig = calculate ( items, numOfItems, &result);
if ( sig == 0) {
printf ( "This is not a valid operation. \n");
} else {
printf ( "The calculation equals %lf \n", result);
}
remove ( "temp");
return 0;
}