I can't assign values to a char array in loop, doing it contains same value in every array value
for example this works
char* foo[3];
foo[0] = "mango"; foo[1] = "kiwi"; foo[2] = "banana";
int i=0; for(i=0;i<3;i++)
{
printf("%s\n",foo[i]);
}
but this doesn't and I don't understand why.
char* foo[3]; int i=0;
for(i=0;i<3;i++) {
char temp[5];
sprintf(temp,"VAL:%d",i);
foo[i] = temp;
}
for(i=0;i<3;i++)
{
printf("%s\n",foo[i]);
}
please help and thanks in advance
The problem in the second snippet is that all elements in foo point to the same variable, temp, which is out of scope when the second for is executed, which is undefined behaviour. Even if it was not out of scope all elements in foo would point to same buffer which is incorrect.
To correct, you need to make a copy of temp and store it in each element of foo. This could be achieved by using strdup() if available (if not malloc() and strcpy()):
for (i = 0; i < sizeof(foo)/sizeof(foo[0]); i++)
{
char temp[6];
snprintf(temp, sizeof(temp), "VAL:%d", i);
foo[i] = strdup(temp); /* Must be free()d later. */
}
or changing the type of foo (as already suggested by unwind):
char foo[3][6];
int i;
for (i = 0; i < sizeof(foo)/sizeof(foo[0]);i++)
{
snprintf(foo[i], sizeof(foo[i]), "VAL:%d", i);
}
Other changes:
temp to 6 as it requires the 5 characters VAL:%d plus the null terminator appended by sprintf(). The posted code has a buffer overrun because of this.snprintf() to avoid buffer overrun.sizeof(foo)/sizeof(foo[0]) to calculate number of elements in array foo instead of hard-coding the element count.You have to remember that in C, a char* does not actually store a string, but it stores the address of a memory location that will be treated as the first character of a string.
In your first example, each element of the foo array holds the address of a different string-literal.
In your second example, each element of the foo array is made to point to the local variable temp. Although each iteration through the loop results in a separate instance of temp, any sane compiler will place all those instances on top of each other, thus giving the results you experience.
The solution is either to use a 2D array:
char foo[3][6]; int i=0;
for (i=0; i<3; i++) { sprintf(foo[i],"VAL:%d", i); }
for (i=0; i<3; i++) { printf("%s\n", foo[i]); }
Or to use dynamic allocation:
char* foo[3]; int i=0;
for (i=0; i<3; i++) { char* temp = malloc(6); sprintf(temp, "VAL:%d", i); foo[i] = temp; }
for (i=0; i<3; i++) { printf("%s\n", foo[i]); free(foo[i]); }
The temp variable in the loop is probably re-using the same memory on each iteration, so you end up writing the same address into each foo[] slot. Then, even worse, that address becomes invalid to access when you exit the loop, causing undefined behavior.
You need to dynamically allocate each string, or use a string array with space for the actual string (char foo[3][20]; or something).
Two problems:
1) You've declared temp as char temp[5] and you are stuffing it with more than 4 characters:
sprintf(temp,"VAL:%d",i);
there by causing sprintf to write beyond the end of the array as it has to write the terminating \0
2) The variable temp is local to the for loop and it goes out of scope once the loop ends.
for(i=0;i<3;i++) { char temp[5]; sprintf(temp,"VAL:%d",i); foo[i] = temp; }assigns a pointer to a local variable that doesn't exist anymore after the loop body finished. Didn't your compiler warn about that?