将不同的值存储在char *向量中，数组中的所有位置都具有相同的值

Question

Let me explain my problem

First of all, imagine I randomly generated the number "5".

I want to convert this integer to a char* so I can display it using my Draw_String() method.

I want to have "5" as the central position which would be [1] in a char* array[2].

In position [0] I want the lower adjacent number "4"

In position [2] I want the higher adjacent number "6"

so

[0] = 4 [1] = 5 [2] = 6

Then I want to display the contents of the vector positions, like this.

4
5
6

However, my output is more like this:

5
5
5

No matter what number I generate, and no matter which operations I perform, I will only get the number "5" in all of my vectors positions.

Code

int startpoint = rand() % 7 + 1;
int assigner = -1;
for (int i = 0; i < 3; i++)
{
    startpoint += assigner;
    convert = std::to_string(startpoint);
    char *converted = &convert[0u];
    blocks.push_back(converted);
    assigner++;
}

//spin around twice for visual effect.

for (int counter = 0; counter < 2; counter++)
{
    Draw_String(drawY, drawX - 1, blocks.at(0));
    Draw_String(drawY, drawX, blocks.at(1));
    Draw_String(drawY, drawX + 1, blocks.at(2));
}

if anyone could help out I'd appreciate it.

Don't ask me to get rid of char* .

Answer 1

Your code has undefined behavior: it harvests a pointer to the internal string buffer from inside the conver string

char *converted = &convert[0u];

but the buffer can become invalid the next time convert string is re-assigned. The behavior that you see suggests that the buffer gets reused, rather than re-allocated, so the same pointer gets stored in the blocks vector.

The best solution would be to store std::string directly, because it would manage both copying and de-allocation for you:

std::vector<std::string> blocks;
...
blocks.push_back(std::to_string(startpoint));

If you insist on using pointers, make a copy of the string before adding it to the container:

char *copy = new char[strlen(converted)+1];
strcpy(copy, converted);
blocks.push_back(copy);

Now you are required to call delete[] on each element of blocks when you are done in order to avoid memory leaks.

Answer 2

If you needto use char* and not std::string than you need to allocate it and copy content of the string intead of copy the pointer.

instead of:

char *converted = &convert[0u];

use:

char *converted = malloc(convert.length() + 1); // number + '\0'
// can also use char *converted = new char[convert.length() + 1];
strcpy(converted, convert.c_str())

Now, no matter what happens to convert, you have control on your converted data. (take care to also delete it when you don't need it)

Otherwise when convert changes, converted can change or point to invalid data. This is because in the original implementation converted is pointing to data that is managed by convert object.