Open the file but do not save the args
-
#include #include #include int main(int *op[]) {
char n;
int total;
int ct = 0, oper1 = 0, oper2 = 0, oper3 = 0, i = 0;
FILE *arq1;arq1 = fopen("Maq1.log", "r"); //open file
while (!feof(arq1)) {
n = fgetc(arq1);
if (n == "\n") { //count lines
ct++;
}
}
op = (int*) calloc(ct, sizeof(int)); //alloc memory
if(op==0) {
printf("Não houve memoria alocada!\n");
return 0;
}
for(i=0;!feof(arq1);i++) {
fscanf(",,,%d,", &op[ct]); //I think this is the error
}
for(i=0;i<=ct;i++) { //do the count
if(op[ct] == 1){
oper1++;
}
if(op[ct] == 2){
oper2++;
}
if(op[ct] == 3){
oper3++;
}
}
total = oper1 + oper2 + oper3;
printf("Operacao 1: %d \n Operacao 2: %d \n Operacao 3: %d \n Total: %d \n", oper1, oper2, oper3, total); //and print!
fclose(arq1);
return 0;
}Can you guys help me on this Code? I have no idea of what is happening, but when I run the code he prints
Operacao 1: 0
Operacao 2: 0
Operacao 3: 0
Total: 0Sorry for the bad english!
-
#include #include #include int main(int *op[]) {
char n;
int total;
int ct = 0, oper1 = 0, oper2 = 0, oper3 = 0, i = 0;
FILE *arq1;arq1 = fopen("Maq1.log", "r"); //open file
while (!feof(arq1)) {
n = fgetc(arq1);
if (n == "\n") { //count lines
ct++;
}
}
op = (int*) calloc(ct, sizeof(int)); //alloc memory
if(op==0) {
printf("Não houve memoria alocada!\n");
return 0;
}
for(i=0;!feof(arq1);i++) {
fscanf(",,,%d,", &op[ct]); //I think this is the error
}
for(i=0;i<=ct;i++) { //do the count
if(op[ct] == 1){
oper1++;
}
if(op[ct] == 2){
oper2++;
}
if(op[ct] == 3){
oper3++;
}
}
total = oper1 + oper2 + oper3;
printf("Operacao 1: %d \n Operacao 2: %d \n Operacao 3: %d \n Total: %d \n", oper1, oper2, oper3, total); //and print!
fclose(arq1);
return 0;
}Can you guys help me on this Code? I have no idea of what is happening, but when I run the code he prints
Operacao 1: 0
Operacao 2: 0
Operacao 3: 0
Total: 0Sorry for the bad english!
Have you tried stepping through the code using a debugger?
"the debugger doesn't tell me anything because this code compiles just fine" - random QA comment "Facebook is where you tell lies to your friends. Twitter is where you tell the truth to strangers." - chriselst "I don't drink any more... then again, I don't drink any less." - Mike Mullikins uncle
-
#include #include #include int main(int *op[]) {
char n;
int total;
int ct = 0, oper1 = 0, oper2 = 0, oper3 = 0, i = 0;
FILE *arq1;arq1 = fopen("Maq1.log", "r"); //open file
while (!feof(arq1)) {
n = fgetc(arq1);
if (n == "\n") { //count lines
ct++;
}
}
op = (int*) calloc(ct, sizeof(int)); //alloc memory
if(op==0) {
printf("Não houve memoria alocada!\n");
return 0;
}
for(i=0;!feof(arq1);i++) {
fscanf(",,,%d,", &op[ct]); //I think this is the error
}
for(i=0;i<=ct;i++) { //do the count
if(op[ct] == 1){
oper1++;
}
if(op[ct] == 2){
oper2++;
}
if(op[ct] == 3){
oper3++;
}
}
total = oper1 + oper2 + oper3;
printf("Operacao 1: %d \n Operacao 2: %d \n Operacao 3: %d \n Total: %d \n", oper1, oper2, oper3, total); //and print!
fclose(arq1);
return 0;
}Can you guys help me on this Code? I have no idea of what is happening, but when I run the code he prints
Operacao 1: 0
Operacao 2: 0
Operacao 3: 0
Total: 0Sorry for the bad english!
The three
if()conditions in yourfor()loop are always looking atop[ct]rather thanop[i]. I do not think that is your intent. ;) A date with the debugger, as jeron1 suggested, would have unveiled that."One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
-
#include #include #include int main(int *op[]) {
char n;
int total;
int ct = 0, oper1 = 0, oper2 = 0, oper3 = 0, i = 0;
FILE *arq1;arq1 = fopen("Maq1.log", "r"); //open file
while (!feof(arq1)) {
n = fgetc(arq1);
if (n == "\n") { //count lines
ct++;
}
}
op = (int*) calloc(ct, sizeof(int)); //alloc memory
if(op==0) {
printf("Não houve memoria alocada!\n");
return 0;
}
for(i=0;!feof(arq1);i++) {
fscanf(",,,%d,", &op[ct]); //I think this is the error
}
for(i=0;i<=ct;i++) { //do the count
if(op[ct] == 1){
oper1++;
}
if(op[ct] == 2){
oper2++;
}
if(op[ct] == 3){
oper3++;
}
}
total = oper1 + oper2 + oper3;
printf("Operacao 1: %d \n Operacao 2: %d \n Operacao 3: %d \n Total: %d \n", oper1, oper2, oper3, total); //and print!
fclose(arq1);
return 0;
}Can you guys help me on this Code? I have no idea of what is happening, but when I run the code he prints
Operacao 1: 0
Operacao 2: 0
Operacao 3: 0
Total: 0Sorry for the bad english!
This has nothing to do with the solution to your problem, as DavidCrow has already given you a clue. I'm just wondering if your compiler isn't giving you any warnings about your definition of main().
int main(int *op[])definitely isn't one of the normal ways to define main() Normally it would be one ofint main()
int main(void)
int main(int argc, char *argv[])
int main(int argc, char **argv)On some systems you might also be able to use
int main(int argc, char *arg[], char *envp[])Note that as a function parameterchar *arg[]andchar **argare equivalent. You also have a type issues withop. It's declared asint *op[], which is an array of pointer to int, but you are using it as a pointer to int. What you probably want to do is:int main(void)
{
...
int \*op;
op = (int\*)calloc(ct, sizeof *op);
...
}If you're wondering why I used
sizeof *opas the second argument tocalloc(), consider what might happen if you decide you want to changeopfrom anintto along, and what might happen if you forget that you need change the call to calloc() too. -
Have you tried stepping through the code using a debugger?
"the debugger doesn't tell me anything because this code compiles just fine" - random QA comment "Facebook is where you tell lies to your friends. Twitter is where you tell the truth to strangers." - chriselst "I don't drink any more... then again, I don't drink any less." - Mike Mullikins uncle
gcc -c "trabalho.c" -o "trabalho.o"
trabalho.c: In function ‘main’:
trabalho.c:22:11: warning: comparison between pointer and integer
if (n == "\n") { //count lines
^
trabalho.c:35:5: warning: format ‘%d’ expects argument of type ‘int *’, but argument 3 has type ‘int’ [-Wformat=]
fscanf(arq1, ",,,%d,", op[ct]); //I think this is the error
^
g++ -o "trabalho.o"
Process terminated with status 0 (0 minute(s), 0 second(s))
0 error(s), 2 warning(s) (0 minute(s), 0 second(s))Checking for existence: /home//
Executing: xterm -T 'dest" (in origem)
Process terminated with status 0 (0 minute(s), 2 second(s))When I compile the .c archive, appears just those 2 warnings
-
The three
if()conditions in yourfor()loop are always looking atop[ct]rather thanop[i]. I do not think that is your intent. ;) A date with the debugger, as jeron1 suggested, would have unveiled that."One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
Thank you! I didn't see that error of logic... --' But still giving me the same answer :/ I think the error is when he count the lines on
char n;
while (!feof(arq1)) {
n = fgetc(arq1);
if (n == "\n") { //count lines
ct++;
}
}warning: assignment makes pointer from integer without a cast and...
for(i=0;!feof(arq1);i++) {
fscanf(arq1, ",,,%d,", op[i]); //I think this is the error
}format ‘%d’ expects argument of type ‘int *’, but argument 3 has type ‘int’ [-Wformat=]|
-
This has nothing to do with the solution to your problem, as DavidCrow has already given you a clue. I'm just wondering if your compiler isn't giving you any warnings about your definition of main().
int main(int *op[])definitely isn't one of the normal ways to define main() Normally it would be one ofint main()
int main(void)
int main(int argc, char *argv[])
int main(int argc, char **argv)On some systems you might also be able to use
int main(int argc, char *arg[], char *envp[])Note that as a function parameterchar *arg[]andchar **argare equivalent. You also have a type issues withop. It's declared asint *op[], which is an array of pointer to int, but you are using it as a pointer to int. What you probably want to do is:int main(void)
{
...
int \*op;
op = (int\*)calloc(ct, sizeof *op);
...
}If you're wondering why I used
sizeof *opas the second argument tocalloc(), consider what might happen if you decide you want to changeopfrom anintto along, and what might happen if you forget that you need change the call to calloc() too. -
Thank you! I didn't see that error of logic... --' But still giving me the same answer :/ I think the error is when he count the lines on
char n;
while (!feof(arq1)) {
n = fgetc(arq1);
if (n == "\n") { //count lines
ct++;
}
}warning: assignment makes pointer from integer without a cast and...
for(i=0;!feof(arq1);i++) {
fscanf(arq1, ",,,%d,", op[i]); //I think this is the error
}format ‘%d’ expects argument of type ‘int *’, but argument 3 has type ‘int’ [-Wformat=]|
honor3us wrote:
if (n == "\n") { //count lines
Why are you not using:
if (n == '\n')
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
-
honor3us wrote:
if (n == "\n") { //count lines
Why are you not using:
if (n == '\n')
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
-
Another error, sorry :/ I changed to 'n' and &op[i] and all the warnings are gonne! :) The programs runs perfectly but all the values of vector op[] are == 0. ex:
printf("%d\n%d", op[4], op[5]);
The answer still
0
0I'm really thankful! :D
See here.
"One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles
-
Thank you! You help me a lot! But the compiler still having problem to count lines and fscanf the data I need, I don't have a clue what is going on--'
You might want to look at this section of the code ... there's definitely an issue here:
for( i=0; i<=ct; i++) { //do the count
if(op[ct] == 1){
oper1++;
}Also, you are accessing past the end of the array *op[]* -- what should the end condition for the loop be?
-
#include #include #include int main(int *op[]) {
char n;
int total;
int ct = 0, oper1 = 0, oper2 = 0, oper3 = 0, i = 0;
FILE *arq1;arq1 = fopen("Maq1.log", "r"); //open file
while (!feof(arq1)) {
n = fgetc(arq1);
if (n == "\n") { //count lines
ct++;
}
}
op = (int*) calloc(ct, sizeof(int)); //alloc memory
if(op==0) {
printf("Não houve memoria alocada!\n");
return 0;
}
for(i=0;!feof(arq1);i++) {
fscanf(",,,%d,", &op[ct]); //I think this is the error
}
for(i=0;i<=ct;i++) { //do the count
if(op[ct] == 1){
oper1++;
}
if(op[ct] == 2){
oper2++;
}
if(op[ct] == 3){
oper3++;
}
}
total = oper1 + oper2 + oper3;
printf("Operacao 1: %d \n Operacao 2: %d \n Operacao 3: %d \n Total: %d \n", oper1, oper2, oper3, total); //and print!
fclose(arq1);
return 0;
}Can you guys help me on this Code? I have no idea of what is happening, but when I run the code he prints
Operacao 1: 0
Operacao 2: 0
Operacao 3: 0
Total: 0Sorry for the bad english!
-
The program is working fine now, just needed to close the file and open again to read the data. Ty for the help everyone!
honor3us wrote:
...just needed to close the file and open again...
Or call
rewind()."One man's wage rise is another man's price increase." - Harold Wilson
"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons
"You can easily judge the character of a man by how he treats those who can do nothing for him." - James D. Miles